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The sum of the currents I1 and I2 in the figure above isThe resistor 10Ω is in...


Question

The sum of the currents I1 and I2 in the figure above is

The resistor 10Ω is in parallel with 5Ω and 10Ω. Therefore, the same potential difference will flow through them.

Note that 5Ω and 10Ω is in series

V = 12V

R = 10Ω

V = IR

12 = 10I2

I2 = 12/10 = 1.2A

5Ω and 10Ω in series

Reff = 5Ω + 10Ω = 15Ω

V = 12V

V = IR

12 = 15I1

I1 = 12/15 = 0.8A

I = I1 + I2 = 1.2A + 0.8A = 2A

Options

A) 0.6A

B) 1.2A

C) 2.0A

D) 0.8A


The correct answer is C.

Explanation:

The sum of the currents I1 and I2 in the figure above is

The resistor 10Ω is in parallel with 5Ω and 10Ω. Therefore, the same potential difference will flow through them.
Note that 5Ω and 10Ω is in series
V = 12V
R = 10Ω
V = IR
12 = 10I2
I2 = 12/10 = 1.2A
5Ω and 10Ω in series
Reff = 5Ω + 10Ω = 15Ω
V = 12V
V = IR
12 = 15I1
I1 = 12/15 = 0.8A
I = I1 + I2 = 1.2A + 0.8A = 2A


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Dicussion (1)

  • The sum of the currents I1 and I2 in the figure above is

    The resistor 10Ω is in parallel with 5Ω and 10Ω. Therefore, the same potential difference will flow through them.
    Note that 5Ω and 10Ω is in series
    V = 12V
    R = 10Ω
    V = IR
    12 = 10I2
    I2 = 12/10 = 1.2A
    5Ω and 10Ω in series
    Reff = 5Ω + 10Ω = 15Ω
    V = 12V
    V = IR
    12 = 15I1
    I1 = 12/15 = 0.8A
    I = I1 + I2 = 1.2A + 0.8A = 2A

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