Question

A battery of emf 24 volt and internal resistance r is connected to a circuit having 2 parallel resistors of 3 ohm and 6 ohm resistor connected in series with 8 ohm resistor.the current flowing in 3 ohm resistor is 0.8 A.calculate internal resistance and terminal pd.

Answer 1

Solution

Emf of battery E=24 V

Resistances

R$_1=3\Omega$

$R_2=6\Omega$

Current in $3\Omega$ resistance =0.8A

Resistances $3 \Omega$ and $6\Omega$ in parallel combination so voltage should be same their end.

$0.8\times3=i\times6$

$i= 0.4A$

Where i is current in $6 \Omega$ resistance.

Therefore total current in circuit $I=0.8+0.4=1.2A$

Now equivalent resistance of circuit

$R=10\Omega$

Therefore Potential difference

(pd)

V$=IR=1.2\times10=12V$

Now

Internal resistance is given by

$r=\frac{E-V}{I}=\frac{24-12}{1.2}=10\Omega$