In the given circuit the cells $E_1$ and $E_2$ have EMFs $5\,\text{V}$ and $9\,\text{V}$ and internal resistances $0.3\,\Omega$ and $1.2\,\Omega$ respectively. Calculate the current flowing through the $3\,\Omega$ resistance.

$I_{3\Omega}=\dfrac{1}{3}\,\text{A}\approx 0.33\,\text{A}$.

CBSE Class 12 · 2006 · Current ElectricityShort Answer

In the given circuit the cells

E_{1}

and

E_{2}

have EMFs

5 V

and

9 V

and internal resistances

0.3 Ω

and

1.2 Ω

respectively. Calculate the current flowing through the

3 Ω

resistance.

Answer: $I_{3Ω} = \frac{1}{3} A \approx 0.33 A$ .

The two cells oppose, so the net EMF is

E_{2} - E_{1} = 9 - 5 = 4 V

. Adding the internal resistances and the series resistor to the parallel pair, the total circuit resistance comes to

8 Ω

, giving the main current

I = \frac{4}{8} = 0.5 A

. This divides between the

6 Ω

and

3 Ω

branches; the part through the

3 Ω

branch is

I_{3Ω} = \frac{6}{6 + 3} \times 0.5 = \frac{1}{3} A .

CBSE Class 12 Board · 2006 · Subjective