Question

7. Applying the point rule at point a in the circuit below, (A) $I_1 + I_2 = -I_3$ (B) $I_1 + I_2 = I_3$ (C) $I_1 + I_3 = I_2$ (D) $I_1 = I_2 + I_3$ 8. Applying the loop rule to the loop a, b, c, d, a (A) $2V + 3\Omega I_1 - 3I_2 - 5V = 0$ (B) $2V - 3\Omega I_1 + 3\Omega I_2 - 5V = 0$ (C) $-2V - 3\Omega I_1 - 3\Omega I_2 + 5V = 0$

          7. Applying the point rule at point a in the circuit
below,
(A) $I_1 + I_2 = -I_3$
(B) $I_1 + I_2 = I_3$
(C) $I_1 + I_3 = I_2$
(D) $I_1 = I_2 + I_3$
8. Applying the loop rule to the loop a, b, c, d, a
(A) $2V + 3\Omega I_1 - 3I_2 - 5V = 0$
(B) $2V - 3\Omega I_1 + 3\Omega I_2 - 5V = 0$
(C) $-2V - 3\Omega I_1 - 3\Omega I_2 + 5V = 0$

7. Applying the point rule at point a in the circuit
below,
(A) I1 + I2 = -I3
(B) I1 + I2 = I3
(C) I1 + I3 = I2
(D) I1 = I2 + I3
8. Applying the loop rule to the loop a, b, c, d, a
(A) 2V + 3Ω I1 - 3I2 - 5V = 0
(B) 2V - 3Ω I1 + 3Ω I2 - 5V = 0
(C) -2V - 3Ω I1 - 3Ω I2 + 5V = 0

Added by Alba M.

College Physics

Eugenia Etkina, Michael Gentle, Alan Van… 1st Edition

Chapter 16

Instant Answer

Solved by Expert Sam Stansfield

08/22/2023

Step 1

00 volts, the equation becomes 1+1v=5.00v. Show more…

Show all steps

Thanks for your feedback!

7. Applying the point rule at point a in the circuit below, (A) I1 + I2 = -I3 (B) I1 + I2 = I3 (C) I1 + I3 = I2 (D) I1 = I2 + I3 8. Applying the loop rule to the loop a, b, c, d, a (A) 2V + 3Î©I1 - 3Î©I2 - 5V = 0 (B) 2V - 3Î©I1 + 3Î©I2 - 5V = 0 (C) -2V - 3Î©I1 - 3Î©I2 + 5V = 0 19 5.00V 30