An electric circuit is given below. V₁ = 1 V and Resistance R = 1000 Ω.
The current through the resistance R is very close to 1 mA and the voltage across point A and B, VAB = 1 V. Now the circuit is changed to :
where value of V₂ = 5 V. The internal resistances of both the batteries are 0.1 Ω. The current through the resistance R is about :
1.0 mA
1.2 mA
3.0 mA
5.0 mA
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-1 – 2024
The second part describes a changed circuit with V2=5V and internal resistances of both batteries (V1 and V2) are 0.1Ω. Assuming A and B are still the points across which R is connected, and both batteries are connected between A and B, the most plausible configuration is that the batteries V1 (1V, 0.1Ω) and V2 (5V, 0.1Ω) are connected in parallel across A and B, and the resistor R (1000Ω) is also connected between A and B.
In a parallel circuit with multiple voltage sources, we can use nodal analysis or source transformation. Using nodal analysis with B as the reference node (0V), the voltage at A is V_A.
Current through R: I_R = V_A / R
Current from V1 source: I_1 = (V1 – V_A) / r1 = (1 – V_A) / 0.1
Current from V2 source: I_2 = (V2 – V_A) / r2 = (5 – V_A) / 0.1
Applying KCL at node A (sum of currents leaving A is zero):
I_R + I_1 + I_2 = 0 is incorrect. The sources are supplying current towards A, so currents entering A should be summed to currents leaving A. Let’s assume currents leave A towards R, and into the positive terminals of V1 and V2. This means V_A is higher than the battery terminals connected to A. The positive terminals are connected to A, and negative terminals to B. So current flows from A through R to B, from A through V1 (r1) to B, and from A through V2 (r2) to B.
Currents leaving A: I_R = V_A / R, I_1 = V_A – V1 / r1? No. This implies V_A is higher than the source terminal.
Let’s redraw with proper node labels. A and B are the terminals. R is between A and B. Battery 1 (V1, r1) is between A and B (positive to A, negative to B). Battery 2 (V2, r2) is between A and B (positive to A, negative to B).
Using superposition or combining parallel sources:
Equivalent voltage source V_eq = (V1/r1 + V2/r2) / (1/r1 + 1/r2)
Equivalent internal resistance r_eq = 1 / (1/r1 + 1/r2)
V_eq = (1V/0.1Ω + 5V/0.1Ω) / (1/0.1Ω + 1/0.1Ω) = (10 A + 50 A) / (10 S + 10 S) = 60 A / 20 S = 3 V.
r_eq = 1 / (10 S + 10 S) = 1 / 20 S = 0.05 Ω.
Now the circuit is a single equivalent source V_eq=3V with internal resistance r_eq=0.05Ω connected across R=1000Ω.
The voltage across R is V_AB = V_eq * R / (R + r_eq) = 3V * 1000Ω / (1000Ω + 0.05Ω) = 3V * 1000 / 1000.05.
V_AB ≈ 3V.
The current through R is I_R = V_AB / R = (3V * 1000 / 1000.05) / 1000Ω = 3V / 1000.05Ω.
I_R ≈ 3 / 1000 = 0.003 A = 3 mA.
Calculating precisely: 3 / 1000.05 ≈ 0.00299985 A ≈ 2.99985 mA.
This is very close to 3.0 mA.
– The second part adds V2=5V and internal resistances 0.1Ω for *both* batteries.
– Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence.
– When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq.
– Formula for parallel sources: V_eq = (Σ Vi/ri) / (Σ 1/ri) and r_eq = 1 / (Σ 1/ri).
– Calculate V_eq and r_eq for V1 (1V, 0.1Ω) and V2 (5V, 0.1Ω).
– Calculate the current through R (1000Ω) when connected across the equivalent source.
– Calculation gives current ≈ 3 mA.