Current Electricity — Question 30

(a) In the circuit, there are three parts. In the first part, resistors of 10 Ω and 40 Ω are connected in parallel. If the equivalent resistance of this part is R'_p then

$\dfrac{1}{R'_p} = \dfrac{1}{10} + \dfrac{1}{40} \\[0.5em] \dfrac{1}{R'_p} = \dfrac{4 + 1}{40} \\[0.5em] \dfrac{1}{R'_p} = \dfrac{5}{40} \\[0.5em] R'_p = \dfrac{40}{5} \\[0.5em] R'_p = 8 Ω \\[0.5em]$

In the second part, resistors of 30 Ω, 20 Ω and 60 Ω are connected in parallel. If the equivalent resistance of this part is R''_p then

$\dfrac{1}{R''_p} = \dfrac{1}{30} + \dfrac{1}{20} + \dfrac{1}{60} \\[0.5em] \dfrac{1}{R''_p} = \dfrac{2 + 3 + 1}{60} \\[0.5em] \dfrac{1}{R''_p} = \dfrac{6}{60} \\[0.5em] R''_p = \dfrac{60}{6} \\[0.5em] R''_p = 10 Ω \\[0.5em]$

In the third part, resistors R'_p and R''_p are connected in series. If the equivalent resistance of this part is R_s then

$R_s = 8 + 10 \\[0.5em] R_s = 18 Ω \\[0.5em]$

Hence, the total resistance of the circuit = 18 Ω

(b) Given,

e.m.f. = 1.8 V

effective resistance of the circuit = 18 Ω

I = ?

From Ohm's law

V= IR

Substituting the values in the formula above, we get,

1.8 = I x 18

⇒ I = $\dfrac{1.8}{18}$

⇒ I = 0.1 A

Hence, the reading of ammeter = 0.1 A