Question

Three resistance,$r_{1}$, are combined in series and this combination is connected to a battery of 12V 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

Answer: (D)

(2.18, 4.36, 5.45)V

Answer 2

:

Let current in the circuit is I. Then total resistance in the circuit

$R = R_{1} + R_{2} + R_{3} = 2 + 4 + 5 = 11$

$\because$ V = IR

$\because I = \frac{V}{R} = \frac{12}{11}A$

The potential drop across $2\Omega$resistance

$V_{1} = IR_{1} = \frac{12}{11} \times 2 = 2.18V$

The potential drop across $4\Omega$resistance

$V_{2} = IR_{2} = \frac{12}{11} \times 4 = 4.36V$

The potential drop across $5\Omega$ resistance

$V_{3} = IR_{3} = \frac{12}{11} \times 5 = 5.45V$

Hence,$(V_{1},V_{2},V_{3}) = (2.18,4.36,5.45)V$