Question

Three resistances $2\Omega ,4\Omega ,5\Omega$ are combines in series and this combination is connected to a battery of $12$V emf and negligible internal resistance. The potential drop across these resistances are.

• A. $(5.45,4.36,2.18)$V
• B. $(2.18,5.45,4.36)$V
• C. $(4.36,2.18,5.45)$V
• D. $(2.18,4.36,5.45)$V

Answer 1

The correct option is D $(2.18,4.36,5.45)$V

Given that three resistors connected in series

$R_1=2\Omega$

$R_2=4\Omega$

$R_3=5\Omega$

Since the resistances are in series, the equivalent resistance will be,

$R_{eq}=R_1+R_2+R_3=11\Omega$

Current through the combination of resistors,

$I=\frac{12}{R_{eq}}=\frac{12}{11}A$

The potential drop across each resistance will be different in a series circuit.

Potential drop through Ist resistor,

$V_1=I\times R_1=\frac{24}{11}=2.18A$

Potential drop through Ist resistor,

$V_2=I\times R_2=\frac{48}{11}=4.36A$

Potential drop through Ist resistor,

$V_3=I\times R_3=\frac{60}{11}=5.45A$