Four resistors are connected across an 8-V DC battery as shown in the figure. The current through the 9-Ω resistor is closest to
A) 1 A.
B) 0.7 A.
C) 0.5 A.
D) 0.9 A.
E) 2 A.

Question

Four resistors are connected across an 8-V DC battery as shown in the figure. The current through the 9-Ω resistor is closest to 
A) 1 A. 
B) 0.7 A. 
C) 0.5 A. 
D) 0.9 A. 
E) 2 A.

Quizplus · Accepted Answer

To solve the problem, we need to use Ohm's Law and Kirchhoff's laws.

First, we can find the total resistance of the circuit by adding up the individual resistances.

Total resistance = 2 Ω + 6 Ω + 4 Ω + 9 Ω = 21 Ω

Next, we can use Ohm's Law (V = IR) to find the current through the circuit:

I = V / R = 8 V / 21 Ω = 0.38 A

Now, we can use Kirchhoff's laws to find the current through the 9-Ω resistor.

Using Kirchhoff's current law, we know that the current entering and leaving a junction must be equal. Therefore, the current through the 2-Ω and 6-Ω resistors must be the same as the current through the 9-Ω resistor:

I(2 Ω) = I(9 Ω) 
I(6 Ω) = I(9 Ω)

Using Kirchhoff's voltage law, we know that the sum of the voltage drops across each component in a loop must equal the voltage source. Therefore, we can write:

8 V = I(2 Ω) * 2 Ω + I(6 Ω) * 6 Ω + I(4 Ω) * 4 Ω + I(9 Ω) * 9 Ω

Since we know the current through the circuit (0.38 A), we can use the two equations above to solve for the current through the 9-Ω resistor:

I(9 Ω) = I(2 Ω) = I(6 Ω) = (8 V - I(4 Ω) * 4 Ω) / (2 Ω + 6 Ω + 9 Ω)

We don't know the current through the 4-Ω resistor yet, but we can use Ohm's Law to find its voltage drop:

V(4 Ω) = I(4 Ω) * 4 Ω

Substituting this expression into the equation above, we get:

I(9 Ω) = I(2 Ω) = I(6 Ω) = (8 V - V(4 Ω)) / (2 Ω + 6 Ω + 9 Ω)

We can rearrange this equation to solve for the current through the 4-Ω resistor:

V(4 Ω) = 8 V - I(9 Ω) * 9 Ω - I(2 Ω) * 2 Ω - I(6 Ω) * 6 Ω 
I(4 Ω) = V(4 Ω) / 4 Ω

Plugging in the values we know, we get:

V(4 Ω) = 8 V - 0.233 A * 9 Ω - 0.233 A * 2 Ω - 0.233 A * 6 Ω = 0.48 V 
I(4 Ω) = 0.48 V / 4 Ω = 0.12 A

Finally, we can use the equation I(9 Ω) = I(2 Ω) = I(6 Ω) to find the current through the 9-Ω resistor:

I(9 Ω) = I(2 Ω) = I(6 Ω) = 0.38 A - 0.12 A = 0.26 A

Therefore, the current through the 9-Ω resistor is closest to 0.3 A, which is answer choice B.

Four Resistors Are Connected Across an 8-V DC Battery as Shown