A cell of internal resistance of 0.2 Ω is connected two cells of resistance 6Ω and 8 Ω joined parallel. There is a current of 0.2 A in the 8 Ω coil. Find the emf of cell.

According to the given information, the electric circuit can be drawn as follows:

Given,

$r = 0.2 \;\Omega$

$R_1 = 8\;\Omega$; $I_1 = 0.2\;A$

$R_2 = 6\;\Omega$

Since $8\;\Omega$ and $6\;\Omega$ are in parallel. So the potential difference across in parallel circuit is equal.

i.e. $I_1\;R_1 = I_2\;R_2$

$I_2 =$$\frac{I_1\; R_1}{R_2} = \frac{0.2 \; * \; 8}{6}$$= 0.27 \; A$

Therefore total current $(I) =  I_1 + I_2 = 0.2 + 0.27 = 0.47\;A$

External Resistance $(R) = R_1||R_2$$= \frac{R_1\;R_2}{R_1 + R_2} =$$\frac{6*8}{6+8}$$= 3.4 \; \Omega$

Now,

ஃ Emf $(E) = IR + Ir = I(R+r) = 0.47(3.4 + 0.2) = 1.7\;V$