Question

The diagram below in Fig., 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.
(b) The reading of ammeter A.

Answer 1

Solving parallel combination:
Given,
Across $XY$,
The resistors $R_1=10\Omega \text{and}{R}_2=40\Omega$ are
in parallel.
Across $AB$,
The resistors $R_3=30\Omega ,R_4=20\Omega ,\text{and}{R}_5=60\Omega$ are in parallel.
Resistance across $XY$:
$R_{XY}=\frac{R_1{R}_2}{R_1+R_2}=\frac{10\times 40}{10+40}=8\Omega$
Resistance across $AB$:
$\frac{1}{R_{AB}}=\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}$
$\frac{1}{R_{AB}}=\frac{1}{30}+\frac{1}{20}+\frac{1}{60}=\frac{2+3+1}{60}$
$R_{AB}=10\Omega$
The total resistance of the circuit:
The resistors $R_{XY}=8\Omega ,R_{AB}=10\Omega$ are in series.
Let $R_{eq}$ be the total resistance.
$R_{eq}=R_{XY}+R_{AB}=8+10=18\Omega$
Ammeter reading
Given,
emf of the cell $𝝴=1.8V$
Equivalent resistance of the circuit $R_{eq}=18\Omega$
Applying Ohm’s law for $R_{eq}=18\Omega$, we et
$𝝴=I{R}_{eq}$
Current through the circuit,
$I=\frac{𝝴}{R_{eq}}=\frac{1.8V}{18\Omega }=0.1A$
Hence, the reading of the ammeter is $0.1A$