For an electric circuit given below, the correct combination of voltage (V) and current (I) is
- (a)V = 900 V; I = 18 A
- (b)V = 300 V; I = 5·5 A
- (c)V = 600 V; I = 1 A
- (d)V = 300 V; I = 2 A
Answer
Why
Correct — B, V = 300 V and I = 5·5 A. The printed circuit puts three resistors — 100 Ω, 200 Ω and 300 Ω — side by side between the same two nodes, so they are in parallel. A voltmeter reads across the 100 Ω branch and an ammeter in that branch shows 3 A. The current I marked leaving on the right is the total.
Start with the branch you have data for. Ohm's law on the 100 Ω resistor:
V = 3 A × 100 Ω = 300 V.
Parallel branches share the same potential difference, so all three have 300 V across them. Now take each branch in turn:
100 Ω → 300/100 = 3 A
200 Ω → 300/200 = 1·5 A
300 Ω → 300/300 = 1 A
The total current is the sum of the branch currents:
I = 3 + 1·5 + 1 = 5·5 A.
Why the others are wrong
- (a)V = 900 V; I = 18 A — This treats the 3 A as flowing through all three resistors and multiplies by the sum 600 Ω, or applies the reading to the wrong branch. The ammeter sits in the 100 Ω branch alone, so V = 3 x 100 = 300 V.
- (c)V = 600 V; I = 1 A — Both figures are wrong. 600 V would need 6 A in the 100 Ω branch, and 1 A is only the current in the 300 Ω branch, not the total leaving the combination.
- (d)V = 300 V; I = 2 A — The voltage is right but the current is not. Having found 300 V, every branch must be worked out and the three added: 3 + 1·5 + 1 = 5·5 A, not 2 A.
Concept
Resistors in parallel all share the same potential difference, while the total current divides between them in inverse proportion to their resistances. Kirchhoff's current law then requires the current entering the combination to equal the sum of the branch currents. One measured branch is therefore enough to fix the whole circuit.
The route through the problem matters more than the arithmetic. The ammeter reading belongs to one branch only, so use it there first to get the shared voltage, then work outwards. Trying to start from the total current, or treating 3 A as the whole, produces the inflated figures in option (a). A useful check at the end: the smallest resistance must carry the largest current, and 3 A through the 100 Ω branch is indeed the biggest of the three.
Key facts
- Parallel branches share the same voltage.
- Branch current = shared voltage / that branch's resistance.
- Total current = sum of the branch currents (Kirchhoff's current law).
- Here V = 3 x 100 = 300 V.
- Branches carry 3 A, 1.5 A and 1 A, totalling 5.5 A.
One measured branch fixes the shared voltage; the rest follows — option (b).
Study next
Common traps
- Treating an ammeter reading in one branch as the total current.
- Assuming the branches share the current rather than the voltage.
- Stopping after finding the voltage without summing the branch currents.
NDA prints a parallel network with one meter reading given — convert that reading into the shared voltage first, then compute every branch and add.
Related PYQs
Consider the following part of an electric circuit: The total electrical resistance in the given part of the electric circuit is
- (a) 15/8 ohm
- (b) 15/7 ohm
- (c) 15 ohm
- (d) 17/3 ohm
Answer(b) 15/7 ohm
The same parallel network handled the other way round — there the branch resistances are collapsed into one equivalent value, here one branch's measured current unlocks the whole circuit.
Practice
- practice — not a real PYQ
Three resistors of 10 Ω, 20 Ω and 20 Ω are connected in parallel across a 20 V supply. The total current drawn is
- (a)1 A
- (b)2 A
- (c)4 A
- (d)6 A
Answer(c) 4 A — branch currents are 2 A, 1 A and 1 A, and they add. - practice — not a real PYQ
In a parallel circuit, the quantity that is the same for every branch is
- (a)the current
- (b)the potential difference
- (c)the resistance
- (d)the power
Answer(b) the potential difference — all branches sit between the same two nodes.