Question

For the circuit shown in the figure, choose the correct options.

• A. The current through $10\Omega$ resistor connected across $NP$ is $0.5\text{A}$
• B. The value of $R_1=50\Omega$
• C. The value of $R=14\Omega$
• D. The potential difference across resistor $R$ is equal to $20\text{V}$

Answer 1

The correct option is C The value of $R=14\Omega$

The given circuit can be redrawn as shown in the figure and we can observe that the $R_1,10\Omega ,20\Omega$ resistors are in parallel connection.


In parallel the current distributes in inverse ratio of resistance.

The current through $20\Omega$ resistor is $1\text{A}$

$\Rightarrow$ Current through $10\Omega$ connected across $P$ and $N$ will be $i=2\text{A}$

[$\because R\propto \frac{1}{i}$ for a parallel combination i.e if the resistance is halved hence the current will get doubled and vice-versa.]

$\Rightarrow \frac{R_1}{20\Omega }=\frac{1\text{A}}{0.5\text{A}}$

$\text{Or,}{R}_1=40\Omega$

The net current supplied by battery will be passing through $MN$,

$i_{\text{net}}=0.5+1+2=3.5\text{A}$

For the potential difference in circuit we can write,

$V_{MN}+V_{NP}=69\text{V}$

$\Rightarrow V_{MN}+(i\times 10)=69$

$\Rightarrow V_{MN}=69-20=49\text{V}$

Applying the ohm's law for resistor $R$,

$V_{MN}=i_{\text{net}}R$

$\Rightarrow R=\frac{49}{3.5}=14\Omega$

$\therefore$Option $\left(c\right)$ is the correct answer.

$\underset{–}{\text{Tip}}$: Whenever the resistors are connected in parallel, always think to apply ohm's law across its end so that we may relate the known current and unknown potential difference in the circuit or the vice-versa.