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

Find the voltage on the resistor 3 $\Omega$ according to Ohm's law: $U_1=I_1\times R_1= 0.8\times 3= 2.4$ V. On the second resistor, the current will be: $I_2=\frac{2.4}{6}=0.4$ A. Since the connection is parallel for two resistors,the total current is $I=I_1+I_2=0.8+0.4=1.2$ A, from Ohm's law for a complete circuit: $I=\frac{ \varepsilon }{R+r}$ we Express the internal resistance $r=\frac{\varepsilon-I\times R}{I}=\frac{24-1.2\times 10}{1.2}=10 \Omega$ .

The voltage on the resistor 8 $\Omega$ is $U_3=I_{total}\times R_3=8\times 1.2=9.6$ V, therefore the total voltage is equal to the sum of the voltage of the first two and the third since they are connected in series $U_{total}=U_1+U_3=2.4+9.6=12$ V.