Question

A resistor of 3 Ω is connected to a battery with an electromotive force (e.m.f) of 6 volts. If the internal resistance of the battery is 20 Ω, calculate the current flowing in the circuit.

a) 0.15 A
b) 0.20 A
c) 0.25 A
d) 0.30 A

Answer 1

The current flowing in the circuit is approximately 0.3023 Amps (A).

Step-by-step explanation:

To calculate the current flowing in the circuit, we can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). In this case, the voltage (V) is the electromotive force (e.m.f) of the battery minus the voltage drop across the internal resistance. So the voltage (V) is 6 volts minus (current (I) multiplied by the internal resistance of 20 ohms). The resistance (R) is the sum of the resistor's resistance (3 ohms) and the internal resistance of the battery (20 ohms). Plugging in the values, we get:

V = 6 - (I * 20)

R = 3 + 20

Now we can substitute these values into Ohm's Law and solve for the current (I):

I = V / R

I = (6 - (I * 20)) / (3 + 20)

Simplifying the equation, we get:

I = 6 - 20I / 23

Multiplying both sides by 23 to remove the fraction, we get:

23I = 138 - 20I

Combining like terms, we get:

43I = 138

Dividing both sides by 43, we get:

I = 138 / 43

I ≈ 0.3023

Therefore, the current flowing in the circuit is approximately 0.3023 Amps (A).