Question

6.00 Ω

www
I₁
ww
9.00 Ω
12.0 Ω
What is the current I₁?
12
E1
Hle
13
27.0 V
HIF
Three resistors and two emf devices are connected in the circuit above. The directions defined as
positive for the three currents are given by the arrows through the resistors (so a current moving
the other way would be defined as negative). The current in the middle branch is I3 = 1.03A.
What is the current I2?
What is the potential difference across emf device &1?

Answer 1

Step-by-step explanation:

To find the current I2, we need to use Kirchhoff's junction rule, which states that the sum of the currents entering a junction must equal the sum of the currents leaving the junction. At the junction between the 12.0 Ω and 6.00 Ω resistors, the current I2 splits into two branches: one with current I3 = 1.03 A and the other with current I1. So we have:

I2 = I1 + I3 = I1 + 1.03 A

To find I1, we can use Ohm's law for the 9.00 Ω resistor:

V = IR

27.0 V = I1(9.00 Ω)

I1 = 3.00 A

Substituting this into the equation for I2, we get:

I2 = 3.00 A + 1.03 A

I2 = 4.03 A

Therefore, the current I2 is 4.03 A.

To find the potential difference across emf device E1, we can use Kirchhoff's loop rule, which states that the sum of the potential differences around any closed loop in a circuit must be zero. We can start at the lower left corner of the circuit and follow a loop around the outside, going through E1, the 12.0 Ω resistor, and the 6.00 Ω resistor:

Starting at the lower left corner of the circuit, we go up through the 6.00 Ω resistor: V = IR = I2(6.00 Ω) = 4.03 A(6.00 Ω) = 24.2 V (since the current I2 flows from right to left through the resistor).

Next, we go through the emf device E1 from the negative (-) terminal to the positive (+) terminal, which adds to the potential difference: V = 27.0 V.

Finally, we go down through the 12.0 Ω resistor: V = IR = I1(12.0 Ω) = 3.00 A(12.0 Ω) = 36.0 V (since the current I1 flows from bottom to top through the resistor).

Adding these potential differences together, we get:

24.2 V + 27.0 V + 36.0 V = 87.2 V

Since this is a closed loop, the sum of the potential differences must be zero. Therefore, the potential difference across emf device E1 is:

E1 = -87.2 V

The negative sign indicates that the emf device is acting as a battery, with the positive (+) terminal at the lower left corner of the circuit and the negative (-) terminal at the upper right corner. So the emf device E1 is providing a potential difference of 87.2 V in the direction from the negative (-) to the positive (+) terminal.