Kirchhoff Current Law at a node states that

sum of currents entering equals sum leaving

Kirchhoff Voltage Law around a closed loop states that

algebraic sum of voltage drops is zero

DC Circuits and Network Theorems MCQs 2026

111 questions with detailed answers · 36 from past papers · 12 quiz batches available

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Page 1 of 1 Questions 1–10 of 111

Q1 Past Paper · PPSC/FPSC/CSS easy

Kirchhoff Current Law at a node states that
1. A sum of voltages is zero
2. B sum of currents entering equals sum leaving
3. C current is same in all branches
4. D power is zero always
💡 Explanation:
KCL follows from conservation of charge.
Q2 hard

Wheatstone bridge P=100, Q=200, R=150 ohm; unknown S at balance is
1. A 75 ohm
2. B 300 ohm
3. C 50 ohm
4. D 450 ohm
💡 Explanation:
100/200 = 150/S → S = 300 ohm.
Q3 Past Paper · PPSC/FPSC/CSS easy

Kirchhoff Voltage Law around a closed loop states that
1. A sum of currents is zero
2. B algebraic sum of voltage drops is zero
3. C sum of resistances is zero
4. D current equals voltage
💡 Explanation:
KVL follows from conservation of energy.
Q4 Past Paper · PPSC/FPSC/CSS easy

Power dissipated in a resistor carrying current I is
1. A I divided by R
2. B R divided by I only
3. C I times R only
4. D I squared times R
💡 Explanation:
P = I²R = VI = V²/R for a resistor.
Q5 Past Paper · PPSC/FPSC/CSS easy

Conductance G of resistance R equals
1. A R
2. B R squared
3. C 1/R
4. D square root of R
💡 Explanation:
G = 1/R in siemens.
Q6 Past Paper · PPSC/FPSC/CSS easy

In a series circuit the current through each element is
1. A inversely proportional to R always
2. B the same
3. C zero in largest R
4. D divided equally by voltage
💡 Explanation:
Single path: same current everywhere.
Q7 Past Paper · PPSC/FPSC/CSS easy

In a parallel circuit the voltage across each branch is
1. A proportional to branch resistance
2. B always zero
3. C the same
4. D sum of branch voltages
💡 Explanation:
Parallel branches share the same potential.
Q8 Past Paper · PPSC/FPSC/CSS medium

A 100 W, 250 V lamp has hot resistance approximately
1. A 2.5 ohm
2. B 25 ohm
3. C 2500 ohm
4. D 625 ohm
💡 Explanation:
R = V²/P = 62500/100 = 625 ohm.
Q9 Past Paper · PPSC/FPSC/CSS medium

A 220 V, 1000 W heater resistance is approximately
1. A 48.4 ohm
2. B 220 ohm
3. C 22 ohm
4. D 484 ohm
💡 Explanation:
R = V²/P = 48400/1000 = 48.4 ohm.
Q10 easy

Balanced Wheatstone bridge galvanometer shows
1. A zero deflection
2. B maximum deflection always
3. C unstable oscillation always
4. D short circuit current
💡 Explanation:
Zero current in galvanometer at balance.
Q11 easy

Energy in a pure resistor is
1. A stored in magnetic field
2. B dissipated as heat
3. C stored as charge permanently
4. D converted to inductance
💡 Explanation:
Resistors dissipate; they do not store energy.
Q12 easy

A practical ammeter should have
1. A very high internal resistance
2. B infinite resistance
3. C same R as voltmeter
4. D very low internal resistance
💡 Explanation:
Low R minimizes loading error.
Q13 easy

Thevenin equivalent of a linear network is
1. A voltage source Vth in series with Rth
2. B current source only
3. C single resistor only
4. D capacitor in series
💡 Explanation:
Thevenin: Vth + series Rth.
Q14 easy

Norton equivalent of a linear network is
1. A voltage source in parallel with R
2. B short circuit only
3. C current source In in parallel with Rn
4. D open circuit only
💡 Explanation:
Norton: In || Rn.
Q15 medium

Thevenin voltage Vth is measured as
1. A open-circuit terminal voltage
2. B short-circuit current
3. C load current at max power
4. D internal drop only
💡 Explanation:
Vth = Voc at terminals.
Q16 medium

Thevenin resistance Rth is found by
1. A adding all resistors always
2. B measuring loaded voltage only
3. C deactivating sources and looking into terminals
4. D multiplying branch currents
💡 Explanation:
Zero sources; compute R at terminals.
Q17 medium

Norton current In equals
1. A short-circuit terminal current
2. B open-circuit voltage
3. C Thevenin resistance
4. D load current always
💡 Explanation:
In = Isc at terminals.
Q18 medium

For maximum power transfer to load, R_L should equal
1. A Rth
2. B zero
3. C twice Rth always
4. D half Rth always
💡 Explanation:
Max power when R_L = Rth.
Q19 hard

Maximum power when Vth=20 V and Rth=5 ohm with matched load is
1. A 20 W
2. B 100 W
3. C 40 W
4. D 10 W
💡 Explanation:
Pmax = V²/(4R) = 400/20 = 20 W.
Q20 hard

At maximum power transfer the efficiency is
1. A 100 percent
2. B 25 percent
3. C 50 percent
4. D 75 percent
💡 Explanation:
Half power in Rth, half in R_L.
Q21 medium

Superposition applies to
1. A saturated magnetic cores only
2. B nonlinear diodes only
3. C linear bilateral networks
4. D single-source circuits only
💡 Explanation:
Response = sum of individual source responses.
Q22 medium

When using superposition, other voltage sources are replaced by
1. A open circuits
2. B their load resistances
3. C short circuits
4. D infinite inductors
💡 Explanation:
Ideal voltage sources → shorts.
Q23 medium

When using superposition, other current sources are replaced by
1. A short circuits
2. B open circuits
3. C Thevenin resistances
4. D matched loads
💡 Explanation:
Ideal current sources → opens.
Q24 medium

Superposition cannot find directly
1. A node voltage from one source
2. B branch current from one source
3. C total power in a resistor with multiple sources
4. D Thevenin voltage
💡 Explanation:
Power is nonlinear in current.
Q25 medium

If Vth=12 V and Rth=4 ohm, Norton current is
1. A 48 A
2. B 0.33 A
3. C 3 A
4. D 8 A
💡 Explanation:
In = Vth/Rth = 3 A.
Q26 medium

Source transformation: V in series with R becomes current source
1. A V times R in series
2. B V/R in parallel with R
3. C R/V in parallel
4. D V in parallel with R only
💡 Explanation:
I = V/R with same R.
Q27 hard

Reciprocity theorem applies to
1. A nonlinear networks only
2. B unilateral devices only
3. C magnetic saturation regions
4. D linear bilateral networks
💡 Explanation:
Interchange excitation and response in linear bilateral nets.
Q28 medium

Mesh analysis uses
1. A KCL at capacitors only
2. B flux linkage only
3. C only Ohm law at one node
4. D KVL around independent loops
💡 Explanation:
Mesh currents + KVL per loop.
Q29 medium

Nodal analysis uses
1. A KVL around every resistor
2. B only Thevenin equivalents
3. C KCL at nodes with voltage unknowns
4. D magnetic flux balance
💡 Explanation:
Node voltages + KCL.
Q30 medium

Independent nodal equations for n nodes number
1. A n
2. B n plus 1
3. C n squared
4. D n minus 1
💡 Explanation:
One reference node at 0 V.
Q31 hard

Independent mesh equations for b branches and n nodes in planar net
1. A n minus 1
2. B b minus n plus 1
3. C b plus n
4. D b times n
💡 Explanation:
Independent loops = b - n + 1.
Q32 medium

A Wheatstone bridge is balanced when
1. A all resistances equal only
2. B galvanometer reads maximum
3. C supply is zero
4. D ratio of adjacent arms on one side equals the other
💡 Explanation:
Balance: P/Q = R/S.
Q33 hard

Delta to wye: each wye resistor equals
1. A sum of two delta only
2. B largest delta only
3. C reciprocal of delta sum
4. D product of two touching delta resistors divided by sum of all three delta
💡 Explanation:
Standard delta-wye transform.
Q34 hard

Three 6 ohm delta resistors convert to each wye arm of
1. A 6 ohm
2. B 18 ohm
3. C 3 ohm
4. D 2 ohm
💡 Explanation:
Ry = 6×6/18 = 2 ohm.
Q35 medium

Three 9 ohm delta to wye gives each arm
1. A 9 ohm
2. B 27 ohm
3. C 1 ohm
4. D 3 ohm
💡 Explanation:
Ry = 81/27 = 3 ohm.
Q36 medium

Voltage divider: R1 and R2 in series across V, voltage on R2 is
1. A V times R1/(R1+R2)
2. B V times R1/R2
3. C V times R2/(R1+R2)
4. D V always
💡 Explanation:
Series divider formula.
Q37 medium

Current divider: R1 parallel R2, total I, current in R1 is
1. A I times R2/(R1+R2)
2. B I times R1/(R1+R2)
3. C I times R1/R2
4. D I always equally
💡 Explanation:
Current splits inversely with R.
Q38 hard

Millman theorem combines
1. A series capacitors only
2. B parallel voltage sources with series resistances
3. C delta networks only
4. D magnetic cores only
💡 Explanation:
Single equivalent for parallel branches.
Q39 hard

Tellegen theorem relates to
1. A power conservation in networks obeying KCL and KVL
2. B resonance only
3. C motor speed only
4. D capacitive reactance only
💡 Explanation:
Topological power relation.
Q40 hard

Star-delta: each delta resistor from wye uses
1. A simple sum of wye arms
2. B product of all three wye
3. C reciprocal sum only
4. D sum of wye pair products divided by opposite wye arm
💡 Explanation:
Standard Y-delta conversion.
Q41 easy

A 6 V source drives 2 ohm load. Load current is
1. A 12 A
2. B 0.3333333333333333 A
3. C 3 A
4. D 8 A
💡 Explanation:
I = V/R = 3 A.
Q42 easy

A 10 V source drives 5 ohm load. Load current is
1. A 50 A
2. B 0.5 A
3. C 15 A
4. D 2 A
💡 Explanation:
I = V/R = 2 A.
Q43 easy

A 24 V source drives 8 ohm load. Load current is
1. A 192 A
2. B 0.3333333333333333 A
3. C 32 A
4. D 3 A
💡 Explanation:
I = V/R = 3 A.
Q44 easy

A 30 V source drives 10 ohm load. Load current is
1. A 300 A
2. B 0.3333333333333333 A
3. C 40 A
4. D 3 A
💡 Explanation:
I = V/R = 3 A.
Q45 easy

A 15 V source drives 5 ohm load. Load current is
1. A 3 A
2. B 75 A
3. C 0.3333333333333333 A
4. D 20 A
💡 Explanation:
I = V/R = 3 A.
Q46 easy

A 20 V source drives 4 ohm load. Load current is
1. A 80 A
2. B 5 A
3. C 0.2 A
4. D 24 A
💡 Explanation:
I = V/R = 5 A.
Q47 easy

A 12 V source drives 6 ohm load. Load current is
1. A 72 A
2. B 0.5 A
3. C 2 A
4. D 18 A
💡 Explanation:
I = V/R = 2 A.
Q48 easy

A 18 V source drives 6 ohm load. Load current is
1. A 108 A
2. B 0.3333333333333333 A
3. C 3 A
4. D 24 A
💡 Explanation:
I = V/R = 3 A.
Q49 easy

A 9 V source drives 3 ohm load. Load current is
1. A 27 A
2. B 3 A
3. C 0.3333333333333333 A
4. D 12 A
💡 Explanation:
I = V/R = 3 A.
Q50 easy

A 16 V source drives 4 ohm load. Load current is
1. A 64 A
2. B 0.25 A
3. C 4 A
4. D 20 A
💡 Explanation:
I = V/R = 4 A.
Q51 medium

4 ohm, 6 ohm, 10 ohm in series across 20 V gives current
1. A 2 A
2. B 0.5 A
3. C 20 A
4. D 1 A
💡 Explanation:
R = 20 ohm; I = 20/20 = 1 A.
Q52 medium

5 ohm, 5 ohm, 10 ohm in series across 20 V gives current
1. A 1 A
2. B 2 A
3. C 0.5 A
4. D 20 A
💡 Explanation:
R = 20 ohm; I = 20/20 = 1 A.
Q53 medium

1 ohm, 2 ohm, 7 ohm in series across 10 V gives current
1. A 2 A
2. B 1 A
3. C 0.5 A
4. D 10 A
💡 Explanation:
R = 10 ohm; I = 10/10 = 1 A.
Q54 Past Paper · PPSC/FPSC/CSS easy

A 20 ohm resistor carries 3 A. The voltage across it is
1. A 6.666666666666667 V
2. B 60 V
3. C 600 V
4. D 0.15 V
💡 Explanation:
V = IR = 20 × 3 = 60 V.
Q55 medium

3 ohm, 3 ohm, 4 ohm in series across 10 V gives current
1. A 1 A
2. B 2 A
3. C 0.5 A
4. D 10 A
💡 Explanation:
R = 10 ohm; I = 10/10 = 1 A.
Q56 medium

Resistors 12, 6, 4 ohm in parallel give
1. A 22 ohm
2. B 2 ohm
3. C 12 ohm
4. D 36 ohm
💡 Explanation:
1/Req = 1/12+1/6+1/4; Req = 2 ohm.
Q57 medium

Resistors 10, 15, 30 ohm in parallel give
1. A 5 ohm
2. B 55 ohm
3. C 10 ohm
4. D 30 ohm
💡 Explanation:
1/Req = 1/10+1/15+1/30; Req = 5 ohm.
Q58 medium

Resistors 8, 8, 8 ohm in parallel give
1. A 24 ohm
2. B 8 ohm
3. C 8/3 ohm
4. D 4 ohm
💡 Explanation:
1/Req = 1/8+1/8+1/8; Req = 8/3 ohm.
Q59 medium

Resistors 20, 10, 20 ohm in parallel give
1. A 5 ohm
2. B 50 ohm
3. C 20 ohm
4. D 60 ohm
💡 Explanation:
1/Req = 1/20+1/10+1/20; Req = 5 ohm.
Q60 medium

Resistors 6, 3, 2 ohm in parallel give
1. A 11 ohm
2. B 1 ohm
3. C 6 ohm
4. D 18 ohm
💡 Explanation:
1/Req = 1/6+1/3+1/2; Req = 1 ohm.
Q61 Past Paper · PPSC/FPSC/CSS easy

A 12 ohm resistor carries 5 A. The voltage across it is
1. A 2.4 V
2. B 60 V
3. C 600 V
4. D 0.4166666666666667 V
💡 Explanation:
V = IR = 12 × 5 = 60 V.
Q62 Past Paper · PPSC/FPSC/CSS easy

Two resistors 10 ohm and 40 ohm in parallel have equivalent resistance of
1. A 50 ohm
2. B 400 ohm
3. C 8 ohm
4. D 0.25 ohm
💡 Explanation:
Parallel: 1/R = 1/10+1/40; R = 8 ohm.
Q63 Past Paper · PPSC/FPSC/CSS easy

Two resistors 20 ohm and 5 ohm in parallel have equivalent resistance of
1. A 25 ohm
2. B 100 ohm
3. C 1.25 ohm
4. D 4 ohm
💡 Explanation:
Parallel: 1/R = 1/20+1/5; R = 4 ohm.
Q64 Past Paper · PPSC/FPSC/CSS easy

Two resistors 6 ohm and 3 ohm in parallel have equivalent resistance of
1. A 2 ohm
2. B 9 ohm
3. C 18 ohm
4. D 1 ohm
💡 Explanation:
Parallel: 1/R = 1/6+1/3; R = 2 ohm.
Q65 Past Paper · PPSC/FPSC/CSS easy

Two resistors 12 ohm and 4 ohm in parallel have equivalent resistance of
1. A 16 ohm
2. B 3 ohm
3. C 48 ohm
4. D 0.75 ohm
💡 Explanation:
Parallel: 1/R = 1/12+1/4; R = 3 ohm.
Q66 Past Paper · PPSC/FPSC/CSS easy

Resistors 4 ohm, 8 ohm connected in series have equivalent resistance of
1. A 6 ohm
2. B 4 ohm
3. C 32 ohm
4. D 12 ohm
💡 Explanation:
Series: R = 4+8 = 12 ohm.
Q67 Past Paper · PPSC/FPSC/CSS easy

A 15 ohm resistor carries 4 A. The voltage across it is
1. A 3.75 V
2. B 600 V
3. C 60 V
4. D 0.26666666666666666 V
💡 Explanation:
V = IR = 15 × 4 = 60 V.
Q68 Past Paper · PPSC/FPSC/CSS easy

A 50 ohm resistor carries 0.5 A. The voltage across it is
1. A 100 V
2. B 25 V
3. C 250 V
4. D 0.01 V
💡 Explanation:
V = IR = 50 × 0.5 = 25 V.
Q69 Past Paper · PPSC/FPSC/CSS easy

A 100 ohm resistor carries 0.2 A. The voltage across it is
1. A 20 V
2. B 500 V
3. C 200 V
4. D 0.002 V
💡 Explanation:
V = IR = 100 × 0.2 = 20 V.
Q70 Past Paper · PPSC/FPSC/CSS easy

A 25 ohm resistor carries 2 A. The voltage across it is
1. A 50 V
2. B 12.5 V
3. C 500 V
4. D 0.08 V
💡 Explanation:
V = IR = 25 × 2 = 50 V.
Q71 Past Paper · PPSC/FPSC/CSS easy

Two resistors 8 ohm and 8 ohm in parallel have equivalent resistance of
1. A 16 ohm
2. B 64 ohm
3. C 4 ohm
4. D 1 ohm
💡 Explanation:
Parallel: 1/R = 1/8+1/8; R = 4 ohm.
Q72 Past Paper · PPSC/FPSC/CSS easy

Resistors 5 ohm, 7 ohm, 3 ohm connected in series have equivalent resistance of
1. A 15 ohm
2. B 7.5 ohm
3. C 5 ohm
4. D 105 ohm
💡 Explanation:
Series: R = 5+7+3 = 15 ohm.
Q73 Past Paper · PPSC/FPSC/CSS easy

Resistors 2 ohm, 2 ohm, 2 ohm connected in series have equivalent resistance of
1. A 6 ohm
2. B 3 ohm
3. C 2 ohm
4. D 8 ohm
💡 Explanation:
Series: R = 2+2+2 = 6 ohm.
Q74 easy

Balanced Wheatstone bridge galvanometer shows
1. A zero deflection
2. B maximum deflection always
3. C unstable oscillation always
4. D short circuit current
💡 Explanation:
Zero current in galvanometer at balance.
Q75 hard

Wheatstone bridge P=100, Q=200, R=150 ohm; unknown S at balance is
1. A 75 ohm
2. B 300 ohm
3. C 50 ohm
4. D 450 ohm
💡 Explanation:
100/200 = 150/S → S = 300 ohm.
Q76 Past Paper · PPSC/FPSC/CSS easy

A 10 ohm resistor carries 2 A. The voltage across it is
1. A 5 V
2. B 200 V
3. C 20 V
4. D 0.2 V
💡 Explanation:
V = IR = 10 × 2 = 20 V.
Q77 Past Paper · PPSC/FPSC/CSS easy

A 5 ohm resistor carries 4 A. The voltage across it is
1. A 1.25 V
2. B 200 V
3. C 0.8 V
4. D 20 V
💡 Explanation:
V = IR = 5 × 4 = 20 V.
Q78 Past Paper · PPSC/FPSC/CSS easy

Kirchhoff Current Law at a node states that
1. A sum of voltages is zero
2. B sum of currents entering equals sum leaving
3. C current is same in all branches
4. D power is zero always
💡 Explanation:
KCL follows from conservation of charge.
Q79 Past Paper · PPSC/FPSC/CSS easy

Kirchhoff Voltage Law around a closed loop states that
1. A sum of currents is zero
2. B algebraic sum of voltage drops is zero
3. C sum of resistances is zero
4. D current equals voltage
💡 Explanation:
KVL follows from conservation of energy.
Q80 Past Paper · PPSC/FPSC/CSS easy

Power dissipated in a resistor carrying current I is
1. A I divided by R
2. B R divided by I only
3. C I times R only
4. D I squared times R
💡 Explanation:
P = I²R = VI = V²/R for a resistor.
Q81 Past Paper · PPSC/FPSC/CSS easy

Conductance G of resistance R equals
1. A R
2. B R squared
3. C 1/R
4. D square root of R
💡 Explanation:
G = 1/R in siemens.
Q82 Past Paper · PPSC/FPSC/CSS easy

In a series circuit the current through each element is
1. A inversely proportional to R always
2. B the same
3. C zero in largest R
4. D divided equally by voltage
💡 Explanation:
Single path: same current everywhere.
Q83 Past Paper · PPSC/FPSC/CSS easy

In a parallel circuit the voltage across each branch is
1. A proportional to branch resistance
2. B always zero
3. C the same
4. D sum of branch voltages
💡 Explanation:
Parallel branches share the same potential.
Q84 Past Paper · PPSC/FPSC/CSS medium

A 100 W, 250 V lamp has hot resistance approximately
1. A 2.5 ohm
2. B 25 ohm
3. C 2500 ohm
4. D 625 ohm
💡 Explanation:
R = V²/P = 62500/100 = 625 ohm.
Q85 Past Paper · PPSC/FPSC/CSS medium

A 220 V, 1000 W heater resistance is approximately
1. A 48.4 ohm
2. B 220 ohm
3. C 22 ohm
4. D 484 ohm
💡 Explanation:
R = V²/P = 48400/1000 = 48.4 ohm.
Q86 easy

Energy in a pure resistor is
1. A stored in magnetic field
2. B dissipated as heat
3. C stored as charge permanently
4. D converted to inductance
💡 Explanation:
Resistors dissipate; they do not store energy.
Q87 easy

A practical ammeter should have
1. A very high internal resistance
2. B infinite resistance
3. C same R as voltmeter
4. D very low internal resistance
💡 Explanation:
Low R minimizes loading error.
Q88 easy

Thevenin equivalent of a linear network is
1. A voltage source Vth in series with Rth
2. B current source only
3. C single resistor only
4. D capacitor in series
💡 Explanation:
Thevenin: Vth + series Rth.
Q89 easy

Norton equivalent of a linear network is
1. A voltage source in parallel with R
2. B short circuit only
3. C current source In in parallel with Rn
4. D open circuit only
💡 Explanation:
Norton: In || Rn.
Q90 medium

Thevenin voltage Vth is measured as
1. A open-circuit terminal voltage
2. B short-circuit current
3. C load current at max power
4. D internal drop only
💡 Explanation:
Vth = Voc at terminals.
Q91 medium

Thevenin resistance Rth is found by
1. A adding all resistors always
2. B measuring loaded voltage only
3. C deactivating sources and looking into terminals
4. D multiplying branch currents
💡 Explanation:
Zero sources; compute R at terminals.
Q92 medium

Norton current In equals
1. A short-circuit terminal current
2. B open-circuit voltage
3. C Thevenin resistance
4. D load current always
💡 Explanation:
In = Isc at terminals.
Q93 medium

For maximum power transfer to load, R_L should equal
1. A Rth
2. B zero
3. C twice Rth always
4. D half Rth always
💡 Explanation:
Max power when R_L = Rth.
Q94 hard

Maximum power when Vth=20 V and Rth=5 ohm with matched load is
1. A 20 W
2. B 100 W
3. C 40 W
4. D 10 W
💡 Explanation:
Pmax = V²/(4R) = 400/20 = 20 W.
Q95 hard

At maximum power transfer the efficiency is
1. A 100 percent
2. B 25 percent
3. C 50 percent
4. D 75 percent
💡 Explanation:
Half power in Rth, half in R_L.
Q96 medium

Superposition applies to
1. A saturated magnetic cores only
2. B nonlinear diodes only
3. C linear bilateral networks
4. D single-source circuits only
💡 Explanation:
Response = sum of individual source responses.
Q97 medium

When using superposition, other voltage sources are replaced by
1. A open circuits
2. B their load resistances
3. C short circuits
4. D infinite inductors
💡 Explanation:
Ideal voltage sources → shorts.
Q98 medium

When using superposition, other current sources are replaced by
1. A short circuits
2. B open circuits
3. C Thevenin resistances
4. D matched loads
💡 Explanation:
Ideal current sources → opens.
Q99 medium

Superposition cannot find directly
1. A node voltage from one source
2. B branch current from one source
3. C total power in a resistor with multiple sources
4. D Thevenin voltage
💡 Explanation:
Power is nonlinear in current.
Q100 medium

If Vth=12 V and Rth=4 ohm, Norton current is
1. A 48 A
2. B 0.33 A
3. C 3 A
4. D 8 A
💡 Explanation:
In = Vth/Rth = 3 A.
Q101 medium

Source transformation: V in series with R becomes current source
1. A V times R in series
2. B V/R in parallel with R
3. C R/V in parallel
4. D V in parallel with R only
💡 Explanation:
I = V/R with same R.
Q102 hard

Reciprocity theorem applies to
1. A nonlinear networks only
2. B unilateral devices only
3. C magnetic saturation regions
4. D linear bilateral networks
💡 Explanation:
Interchange excitation and response in linear bilateral nets.
Q103 medium

Mesh analysis uses
1. A KCL at capacitors only
2. B flux linkage only
3. C only Ohm law at one node
4. D KVL around independent loops
💡 Explanation:
Mesh currents + KVL per loop.
Q104 medium

Nodal analysis uses
1. A KVL around every resistor
2. B only Thevenin equivalents
3. C KCL at nodes with voltage unknowns
4. D magnetic flux balance
💡 Explanation:
Node voltages + KCL.
Q105 medium

Independent nodal equations for n nodes number
1. A n
2. B n plus 1
3. C n squared
4. D n minus 1
💡 Explanation:
One reference node at 0 V.
Q106 hard

Independent mesh equations for b branches and n nodes in planar net
1. A n minus 1
2. B b minus n plus 1
3. C b plus n
4. D b times n
💡 Explanation:
Independent loops = b - n + 1.
Q107 medium

A Wheatstone bridge is balanced when
1. A all resistances equal only
2. B galvanometer reads maximum
3. C supply is zero
4. D ratio of adjacent arms on one side equals the other
💡 Explanation:
Balance: P/Q = R/S.
Q108 Past Paper · PPSC/FPSC/CSS easy

A 4 ohm resistor carries 7 A. The voltage across it is
1. A 0.5714285714285714 V
2. B 28 V
3. C 280 V
4. D 1.75 V
💡 Explanation:
V = IR = 4 × 7 = 28 V.
Q109 Past Paper · PPSC/FPSC/CSS easy

Resistors 10 ohm, 15 ohm connected in series have equivalent resistance of
1. A 25 ohm
2. B 12.5 ohm
3. C 10 ohm
4. D 150 ohm
💡 Explanation:
Series: R = 10+15 = 25 ohm.
Q110 Past Paper · PPSC/FPSC/CSS easy

Resistors 6 ohm, 3 ohm connected in series have equivalent resistance of
1. A 4.5 ohm
2. B 9 ohm
3. C 6 ohm
4. D 18 ohm
💡 Explanation:
Series: R = 6+3 = 9 ohm.
Q111 Past Paper · PPSC/FPSC/CSS easy

A 8 ohm resistor carries 2.5 A. The voltage across it is
1. A 3.2 V
2. B 200 V
3. C 0.3125 V
4. D 20 V
💡 Explanation:
V = IR = 8 × 2.5 = 20 V.

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