Plss solve 4th qn

Dear Student,

Let the current flowing through the circuit be i and the emf of the battery be E .

Applying KVL in the circuit

E $-$ 20i $-$3i = 0
$\Rightarrow$E = 23 i

According to the question

The voltage across the 20 Ω resistance is 10 V
$\Rightarrow$10 = 20 i
$\Rightarrow$ i = $\frac{1}{2} \mathrm{A}$
So , the emf of the battery is E = 23 $\times \frac{1}{2}=11.5 \mathrm{V}$

When the resistance of 30 ​Ω is connected in the circuit 


Total resistance in the circuit = 20 + 30 + 3 = 53 ​ Ω

Current flowing through the circuit is found by

 $$\begin{gathered} i=\frac{E}{R}= \frac{11.5}{53}=0.216 \mathrm{A} \\ \mathrm{The} \mathrm{potential} \mathrm{drop} \mathrm{across} \mathrm{the} \mathrm{combination} \mathrm{of} \mathrm{the} \mathrm{resistors} \mathrm{will} \mathrm{be} \\ V=i(20+30) \\ V=50\times 0.216=10.84 \mathrm{V} \end{gathered}$$
Regards