Two identical batteries, each with an emf of 18 V and an internal resistance of 1 $\Omega$, are wired in parallel by connecting their positive terminals together and connecting their negative terminals together.The combination is then wired across a 4-$\Omega$ resistor.The potential difference across the 4-$\Omega$ resistor is:
A)4.0 V
B)8.0 V
C)14 V
D)16 V
E)29 V

Question

Quizplus · Accepted Answer

When two identical batteries are connected in parallel, the effective emf remains the same as a single battery (18 V) because the voltage across each battery is the same. The internal resistance of the parallel combination is halved (0.5 Ω) because the resistances are identical. Using Ohm's law and the total resistance in the circuit (which is the sum of the internal resistance of the batteries in parallel and the external resistance, 0.5 Ω + 4 Ω = 4.5 Ω), the total current can be calculated as $I = \frac{V}{R} = \frac{18V}{4.5\Omega} = 4A$. The potential difference across the 4 Ω resistor is then $V = IR = 4A \times 4\Omega = 16V$.

Two Identical Batteries, Each with an Emf of 18 V Ω\OmegaΩ

Two Identical Batteries, Each with an Emf of 18 V $\Omega$