The diagram below in figure shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8 V. Calculate —

(a) the total resistance of the circuit, and

(b) the reading of ammeter A.

(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] Rs = 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,

$1.8 = I \times 18 \\[0.5em] \Rightarrow I = \dfrac{1.8}{18} \\[0.5em] \Rightarrow I = 0.1 A$

Hence, the reading of ammeter = 0.1 A