Question

In the given circuit, the voltmeter reading is 4.5 V. Assuming that the voltmeter is ideal, current through 12Ω resistance is –

• A. 1 A
• B. 0.5 A
• C. 0.25 A
• D. 0.1 A

Answer 1

Answer: (B)

0.5 A

Answer 2

The voltmeter is ideal, its resistance R_v → ∞ Fig. shows the current distribution in the circuit . Voltmeter will not draw any current.

Potential difference across 9 Ω resistance = 4.5 V (given) Hence, current in 9Ω resistance

= $\frac{4.5}{9}$ = 0.5 A (I = $\frac{V}{R}$)

i.e., $I_{1}^{'}$ = 0.5 A

The same current ($I_{1}^{'}$) passes through 3Ω. Obviously, 9 Ω and 3 Ω are in series and their equivalent, i.e., 12Ω is in parallel with 6 Ω between A and B. Dividing the current in the inverse ratio of resistances between A and B,

$\frac{I_{1}^{'}}{I_{1}^{''}}$ = $\frac{6}{12}$ = $\frac{1}{2}$

$I_{1}^{''}$ = 2$I_{1}^{'}$ = 2 × 0.5 = 1 A

and I₁ = $I_{1}^{'}$ + $I_{1}^{''}$ = 0.5 + 1 = 1.5 A

at junction C, I₁ divides into three parts. Since the resistances 10Ω, 12Ω, 15 Ω are in parallel between C and D, current will distribute in the inverse ratio of resistances.

∴ $I_{2}^{'}$ : $I_{2}^{''}$ : $I_{2}^{'''}$ = $\frac{1}{10}$ : $\frac{1}{12}$ : $\frac{1}{15}$ = 6 : 5 : 4

$I_{2}^{'}$ = $\frac{5}{15}$× 1.5 = 0.5 amp

So $I_{2}^{'}$ = 6k, $I_{2}^{''}$ = 5k, $I_{2}^{'''}$ = 4k

(k being a constant of proportionality)

and I₁ = $I_{2}^{'}$ + $I_{2}^{''}$ + $I_{2}^{'''}$ = 15k

but I₁ = 1.5 A

∴ 15 k = 1.5

or k = 0.1

so $I_{2}^{''}$ = 5k = 0.5 A

Thus current through 12Ω resistance is 0.5 A