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

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)

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