Question

A battery of 10 volts emf and points internal resistance of 0.50 ohms is connected in parallel with another battery of 12 volts and internal resistance of 0.80 ohms. The terminals are connected to an external load resistance of 20 ohms. What is the load current?

Answer 1

Given:

$E_1=10\:\rm V$

$r_1=0.50\:\Omega$

$E_2=12\:\rm V$

$r_2=0.80\:\Omega$

$R=20\:\Omega$

The Kirchhoff rules give

$$I_3=I_1+I_2$$

$$-I_1r_1-I_3R+E_1=0$$

$$I_2r_2+I_3R-E_2=0$$

Solution:

$$I_3=\frac{E_1r_2+E_2r_1}{r_1r_2+R(r_1+r_2)}$$

$$I_3=\frac{10*0.80+12*0.50}{0.80*0.50+20*(0.80+0.50)}=0.53\:\rm A$$