1 point 40 The diagram below shows an electrical circuit made up of a cell and three loads. The
voltage established by the cell and the resistance of each load is marked on the
diagram. Select the current flowing through the 3-ohm load.
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1 ohm
12V
2 ohms
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3 ohms
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A. 4 amps
B. 2 amps
C. 5 amps
D. 0.25 amps
SPS106 DOK 3

Option B. 2 A.

Explanation:

The following data were obtained from the question:

Resistance 1 (R1) = 1 Ω

Resistance 2 (R2) = 2 Ω

Resistance 3 (R3) = 3 Ω

Potential difference (V) = 12 V.

Current through the 3 Ω (I3) =?

Next, we shall determine the total resistance of the circuit.

This can be obtained as follow:

Resistance 1 (R1) = 1 Ω

Resistance 2 (R2) = 2 Ω

Resistance 3 (R3) = 3 Ω

Total resistance (R) =.?

R = R1 + R2 + R3

R = 1 + 2 + 3

R = 6 Ω

Next, we shall determine the current flowing in the circuit.

This can be obtained as follow:

From ohm's law,

Voltage (V) = current (I) x resistance (R)

V = IR

Potential difference (V) = 12 V.

Resistance (R) = 6 Ω

Current (I) =?

V = IR

12 = I x 6

Divide both side by 6

I = 12/6

I = 2 A.

Since the circuit is in series connection, therefore, the same current will flow through each resistor.

Therefore, the current through the the 3 Ω load is 2 A.