Question

In a circuit shown, voltmeter reads $3V$ and the ammeter reads $2A$. The emf $E$ (in $Volt$) is

Answer 1

Step 1: Voltage drop across $2.5\Omega$

It is given that reading of ammeter is $2A$.

$\therefore$ Current flowing through $2.5\Omega$ ,

$I_{2.5\Omega }=2A$

Voltage drop across $2.5\Omega$ ,

$V_1=2\times 2.5=5V$


Step 2: Voltage drop across $2\Omega$

Current flowing through $2\Omega$ ,

$I_{2\Omega }=2A$

$\therefore$ Voltage drop across $2\Omega$ ,

$V_2=2\times 2=4V$


Step 3: Apply KVL in the whole circuit

By KVL,

${\sum }_{i=1}^{4}V=0$

$\Rightarrow V_1+V_2+V_3+V_4=0$

Where $V_3$ is the reading of voltmeter, $V_4$ is the emf $E$ and $V_1,V_2$ are as mentioned above.

Now $V_1=5V,V_2=4V$ (Proved above)

Given,

$V_3=3V,V_4=E$

$\Rightarrow -E+5+4+3=0$

$\Rightarrow E=12V$

Hence, EMF of the cell (E) is $12V$.