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luda_lava [24]
3 years ago
15

Determine the current in the 7-ohm resistor for the circuit shown in the figure. Assume that the batteries are ideal and that al

l numbers are accurate to two significant figures. Also determine the current for the 8-ohm and the 4-ohm resistors. Please show your work

Physics
1 answer:
seraphim [82]3 years ago
4 0

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:

<u>I₁ = 1.6 A (through 7 Ohm Resistor)</u>

<u>I₂ = 1.3 A (through 8 Ohm Resistor)</u>

<u>I₃ = I₁ - I₂ = 1.6 A - 1.3 A = 0.3 A (through 4 Ohm Resistor)</u>

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