Question

Three resistors of resistances 3$\left( e = 1.6 \times 10^{- 19}C \right)$, 4$4\Omega$ and 5$2 \times 10^{- 2}\Omega$ are combined in parallel. This combination is connected to a battery of emf 12V and negligible internal resistance, current through each resistor in ampere is

• A. 4, 3, 2.4
• B. 8, 7, 3.4
• C. 2, 5, 1.8
• D. 5, 5, 8.2

Answer 1

Answer: (A)

4, 3, 2.4

Answer 2

: Since the voltage across the circuit is constant. Then currect through $3\Omega$resistor,

$I_{1} = \frac{V}{R_{1}} = \frac{12}{3} = 4A$

The current through $4\Omega$resistor

$I_{2} = \frac{V}{R_{2}} = \frac{12}{4} = 3A$

and the current through $5\Omega$resistor, $I_{3} = \frac{V}{R_{3}} = \frac{12}{5}$

$= 2.4A$