Question

Three resistors of 6.0 ohm, 2.0 ohm and 4.0 ohm are joined to an ammeter A and a cell of emf 6.0 V as shown in figure. Calculate:
(a) the effective resistance of the circuit.
(b) the reading of ammeter.

Answer 1

Simplifying series branch:
Given,
The resistors $R_3=4\Omega \text{and}{R}_2=2\Omega$ are in series.
Resistance of branch:
$R_s=R_2+R_3=2+4=6\Omega$
(a) The effective resistance of the circuit:
The resistors $R_S=6\Omega \text{and}{R}_1=6\Omega$ are in
parallel.
Let $R_{eq}$ be the equivalent resistance,
$R_{eq}=\frac{R_s{R}_1}{R_s+R_1}=\frac{6\times 6}{6+6}=3\Omega$
(b) Ammeter reading
Given,
emf of the cell $𝝴=6V$
Equivalent resistance of the circuit
$R_{eq}=3\Omega$
Applying Ohm’s law for $R_{eq}=3\Omega$, we get
$𝝴=I{R}_{eq}$
Current through the resistor $R_{eq}$ ,
$I=\frac{𝝴}{R_{eq}}=\frac{6V}{3\Omega }=2A$
Hence, the reading of the ammeter is $2A$.