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 current in the 4- $\Omega$resistor is:
A) 1.0 A
B) 2.0 A
C) 4.0 A
D) 3.6 A
E) 7.2 A

Question

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 current in the 4- $\Omega$resistor is:
A) 1.0 A
B) 2.0 A
C) 4.0 A
D) 3.6 A
E) 7.2 A

Quizplus · Accepted Answer

First, we can use the formula for equivalent resistance for resistors in parallel:
$$R_{eq} = \frac{R_1R_2}{R_1+R_2}$$
In this case, the two internal resistances are the same, so we can simplify to:
$$R_{eq} = \frac{1\Omega \cdot 1\Omega}{1\Omega + 1\Omega} = 0.5\Omega$$
Next, we can use the equation for the total emf of batteries in parallel:
$$E_{total} = E_1 = E_2$$
So, the total emf is 18V.
Using Ohm's Law, we can find the current in the circuit:
$$I = \frac{E_{total}}{R_{eq}+R} = \frac{18V}{0.5\Omega + 4\Omega} = \frac{18V}{4.5\Omega} = 4A$$
Therefore, the current in the 4Ω resistor is 4A, so the answer is C.

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

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