Question

Determine the current in the 7-ohm resistor for the circuit shown in the figure. Assume that the batteries are ideal and that all numbers are accurate to two significant figures. Also determine the current for the 8-ohm and the 4-ohm resistors. Please show your work

Answer 1

Answer:

I₁ = 1.6 A (through 7 Ohm Resistor)

I₂ = 1.3 A (through 8 Ohm Resistor)

I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)

Explanation:

Here we consider two loops doe applying Kirchhoff's Voltage Law (KVL). The 1st loop is the left side one with a voltage source of 12 V and the 2nd Loop is the right side one with a voltage source of 9 V. We name the sources and resistor's as follows:

R₁ = 7 Ω

R₂ = 4 Ω

R₃ = 8 Ω

V₁ = 12 V

V₂ = 9 V

Now, we apply KVL to 1st Loop:

V₁ = I₁R₁ + (I₁ - I₂)R₂

12 = 7I₁ + (I₁ - I₂)(4)

12 = 7I₁ + 4I₁ - 4I₂

I₁ = (12 + 4 I₂)/11   ------------ equation (1)

Now, we apply KVL to 2nd Loop:

V₂ = (I₂ - I₁)R₂ + I₂R₃

9 = (I₂ - I₁)(4) + 8I₂

9 = 4I₂ - 4I₁ + 8I₂

9 = 12I₂ - 4I₁   -------------- equation (2)

using equation (1)

9 = 12I₂ - 4[(12 + 4 I₂)/11]

99 = 132 I₂ -  48 - 16 I₂

147 = 116 I₂

I₂ = 147/116

I₂ = 1.3 A

use this value in equation 2:

9 = 12(1.3 A) - 4I₁

4I₁ = 15.6 - 9

I₁ = 6.6 A/4

I₁ = 1.6 A

Hence, the currents through all resistors are:

I₁ = 1.6 A (through 7 Ohm Resistor)

I₂ = 1.3 A (through 8 Ohm Resistor)

I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)

Answer 2

a. The current in the 7-ohm resistor for the given circuit is 1.6 Amperes.

b. The current in the 8-ohm resistor for the given circuit is 1.3 Amperes.

c. The current in the 4-ohm resistor for the given circuit is 0.3 Amperes.

Given the following data:

• Resistor 1 ($R_1$) = 7 Ohms

• Resistor 1 ($R_2$) = 4 Ohms

• Resistor 1 ($R_3$) = 8 Ohms

• Voltage 1 ($V_1$) = 12 Volts

• Voltage 1 ($V_2$) = 9 Volts

To determine the current in the 7-ohm resistor for the given circuit, we would apply Kirchhoff's Voltage Law (KVL):

Applying KVL to the 1st loop:

$V_1 = I_1R_1 + (I_1 - I_2)R_2$

Substituting the given parameters into the formula, we have;

$12 = 7I_1 + (I_1 - I_2)4\\\\12 = 7I_1+4I_1 - 4I_2\\\\12 = 11I_1- 4I_2\\\\11I_1 = 12 + 4I_2\\\\I_1 = \frac{12 \;+\; 4I_2}{11}$...equation 1.

Applying KVL to the 2nd loop:

$V_2 = (I_2 - I_1)R_2 + I_2R_2\\\\9 = (I_2 - I_1)4 + 8I_2\\\\9=4I_2 - 4I_1+8I_2\\\\9=12I_2 - 4I_1$.....equation 2.

Substituting the value of $I_1$ into eqn 2, we would solve for the current through the 8-ohm resistor:

$9=12I_2 -4(\frac{12 \;+\; 4I_2}{11})\\\\9=12I_2 - (\frac{48 \;+\; 16I_2}{11})\\\\99 = 132I_2 -48 -16I_2\\\\99+48=132I_2 -48 -16I_2\\\\147 = 116I_2\\\\I_2=\frac{147}{116} \\\\I_2 = 1.3 \;Amps$

Next, we would solve for $I_1$:

$I_1 = \frac{12 \;+\; 4I_2}{11} \\\\I_1 = \frac{12 \;+\; 4\times 1.3}{11} \\\\I_1 = \frac{12 \;+\; 5.2}{11}\\\\I_1 = \frac{17.2}{11}\\\\I_1 =1.6\;Amps$

For the the 4-ohm resistor:

$I_3 = I_1-I_2\\\\I_3 = 1.6 -1.3\\\\I_3 =0.3\;Amp$

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