Question

Open circuit voltage of a source is 7.86 V and its short circuit current is 9.25 A. Find the current when an external resistance of $2.4\Omega$ is connected

• A. 1.4A
• B. 1.82A
• C. 2.01A
• D. 2.4A

Answer 1

Answer: (D)

2.4A

Open circuit voltage of a source is 7.86 V and its short circuit current is 9.25 A. Find the current when an external resistance of 2.4 is connected
Given
The open circuit voltage is $V_{oc}=7.86V$
The short circuited current is $I_{sc}=9.25A$
The external resistance in the circuit is $=R=2.4\Omega$
The open circuit voltage is the emf of the battery when it is not connected in the circuit and current flowing through battery is short circuit current, given by
$I_{sc}=\frac{E}{r}$

where, $r$ is the internal resistance of the battery.
$r=\frac{E}{I_{sc}}$

$r=\frac{7.86}{9.25}$

$r=0.849\approx 0.85\Omega$

When the battery is connected in circuit with resistance $R$, the current $I$(say), is
$I=\frac{E}{R+r}$

$I=\frac{7.86}{2.4+0.85}$

$I=\frac{7.86}{3.25}$

$I=2.418\approx 2.4A$