Current Electricity — Question 29

(a) In the circuit, there are two parts. In the first part, resistors of 2.0 and 4.0 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = 2 + 4 = 6 Ω

In the second part, R_s = 6.0 and resistor of 6.0 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2}{6} \\[0.5em] R_p = \dfrac{6}{2} \\[0.5em] R_p = 3.0 Ω \\[0.5em]$

Hence, the effective resistance of the circuit = 3 Ω

(b) The reading of ammeter = ?

R = 3 Ω

V = 6.0 V

Using Ohm's law,

V = IR

Substituting the values in the formula above we get,

6 = I x 3
⇒ I = 6 / 3 = 2 A

Hence, the reading of ammeter = 2 A