what is the current in each branch of the circuit? 8 ω: a 10 ω: a 12 ω: a 60.0 v 8.0 ω 10.0 ω 12.0 ω

Explanation:

Step1: Apply Ohm's law ($I = \frac{V}{R}$) for each resistor

For the 8 - Ω resistor, $I_1=\frac{V}{R_1}$

Step2: Substitute values

Given $V = 60.0$ V and $R_1=8.0$ Ω, so $I_1=\frac{60.0}{8.0}=7.5$ A

Step3: For the 10 - Ω resistor

$I_2=\frac{V}{R_2}$, substituting $V = 60.0$ V and $R_2 = 10.0$ Ω, we get $I_2=\frac{60.0}{10.0}=6.0$ A

Step4: For the 12 - Ω resistor

$I_3=\frac{V}{R_3}$, substituting $V = 60.0$ V and $R_3=12.0$ Ω, we get $I_3=\frac{60.0}{12.0}=5.0$ A

Answer:

8 Ω: 7.5 A 10 Ω: 6.0 A 12 Ω: 5.0 A