In the given figure, the emf of the cell is 2.2 V and if internal resistance is \(0.6 \Omega\). Calculate the power dissipated in the whole circuit :

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Current Electricity

In the given figure, the emf of the cell is 2.2 V and if internal resistance is

0.6Ω

. Calculate the power dissipated in the whole circuit :

Answer: (A)

The two vertical sides of the rectangle are plain wires, so all four branches run from the same node

A

to the same node

B

: the

4 Ω

top edge, the

8 Ω

bottom edge, one diagonal of

4 + 8 = 12 Ω

and the other diagonal of

2 + 4 = 6 Ω

, the diagonals crossing without contact. Hence

\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{6} = \frac{6 + 3 + 2 + 4}{24} = \frac{5}{8} \Rightarrow R = 1.6 Ω

Adding the internal resistance, the total is

1.6 + 0.6 = 2.2 Ω

, so

I = \frac{2.2}{2.2} = 1 A

The power dissipated in the whole circuit, internal resistance included, is

P = ε I = 2.2 \times 1 = 2.2 W

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option