Question

Three resistors are connected in series across a 12 V battery. the first resistor has value of 1 $\Omega$, second has a voltage drop of 4 V and third has a power dissipation of 12 W. Then, the value of the circuit current is

• A. 2 A
• B. 6 A
• C. either 2 A or 6 A
• D. none of the above

Answer 1

The correct option is C either 2 A or 6 A

Let, $R_1$, $R_2$ and $R_3$ be the three resistors connected in series across a 12 V battery. $I$ be the current in circuit.

$V=IR$

$V=I(R_1+R_2+R_3)$

$V=I{R}_1+I{R}_2+I{R}_3$ ..................$\left(1\right)$

Now, given that,

$R_1=1\Omega$

$I{R}_2=4V$ and

$P_3=I^2{R}_3\Rightarrow R_3=\frac{P_3}{I^2},P_3=12W$ and $V=12V$.

Using this values in $\left(1\right)$, we get,

$12=I\left(1\right)+4+I\left(\frac{12}{I^2}\right)$

$8=I+\frac{12}{I}$

$I^2+12=8I$

$I^2-8I+12=0$

Solving for roots we get

$I-2=0$ or $I-6=0$

$I=2A$ or $I=6A$