Question

In the circuit in figure, we have $R_2=20\Omega$ , $R_3=15\Omega$, and the current flowing through resistor $R_2$ is 0.3 A. The ammeter shows 0.8 A. Find the resistance $R_1$ :

• A. $20\Omega$
• B. $30\Omega$
• C. $40\Omega$
• D. $60\Omega$

Answer 1

Answer: (D)

Potential drop across $R_2$ = Potential drop across $R_1$ = Potential drop across $R_3$
Considering $R_2$ and $R_3$
$i_2{R}_2=i_3{R}_3$
$0.3\times 20=15\times i_3$
$i_3=0.4A$
Current passing through ammeter $=i_1+i_2+i_3=0.8$
$i_1+0.3+0.4=0.8$
$i_1=0.1A$
PD across $R_1$ should be equal to the potential drop across $R_2$
$i_1{R}_1=i_2{R}_2$
$0.1R_1=0.3\times 20$
$R_1=60ohm$
Hence option D is correct.