The emfs and resistances in the given circuit have the following values.
$E_1=4.2 ~V, E_2=1.9 ~V$
$r_1=2.0 \Omega, r_2=1.6 \Omega, R=6.0 \Omega$
What is the current in the circuit?
Answer & explanation
Correct answer: option 3
Applying Kirchoff's Lawin the circuit
$ E_1 - E_2 -I( R + r_1 + r_2) = 0$
$ I = \frac{E_1-E_2}{R + r_1 + r_2} = \frac{4.2 - 1.9}{6+2+1.6} =\frac{2.3}{9.6} = 240mA$