Question

5. Find $i_w$, $i_x$, $i_y$, and the power for the dependent source. + 50V - 10V - 6. For the circuit shown, find $i_A$, $i_B$, $i_C$, $i_D$, $i_E$, and the power for each of the sources. 6V 3? 5? 3A? 5? 2k? W 3k? 3k? W 2k? 1k? W 1k? W 8k? W 0.5$i_w$ 4k? W 2.5? W 5.5? 1.5A 8? W 30V + 2? W 2? W 3? W 4? W

          5. Find $i_w$, $i_x$, $i_y$, and the power for the dependent source.
+ 
50V
- 
10V
- 
6. For the circuit shown, find $i_A$, $i_B$, $i_C$, $i_D$, $i_E$, and the power for each of the sources.
6V
3?
5?
3A?
5?
2k?
W
3k?
3k?
W
2k?
1k?
W
1k?
W
8k?
W
0.5$i_w$
4k?
W
2.5?
W
5.5?
1.5A
8?
W
30V
+
2?
W
2?
W
3?
W
4?
W

5. Find iw, ix, iy, and the power for the dependent source.
+
50V
-
10V
-
6. For the circuit shown, find iA, iB, iC, iD, iE, and the power for each of the sources.
6V
3?
5?
3A?
5?
2k?
W
3k?
3k?
W
2k?
1k?
W
1k?
W
8k?
W
0.5iw
4k?
W
2.5?
W
5.5?
1.5A
8?
W
30V
+
2?
W
2?
W
3?
W
4?
W

Added by Walter M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sikandar Baig

Step 1

We can do this by combining resistors in series and parallel. First, let's combine the 3Ω and 3Ω resistors in parallel: 1/(1/3 + 1/3) = 1/(2/3) = 3/2 = 1.5Ω Now, let's combine the 1.5Ω and 3Ω resistors in series: 1.5 + 3 = 4.5Ω Now, we can use Ohm's Law to find Show more…

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5. Find ix and the power for the dependent source 2k Î© 3k Î© 3k Î© 2k Î© 1k Î© 1k Î© 2 Î© 2 Î© 50V 10V 8k Î© 4k Î© 0.5A 6. For the circuit shown, find IA, le, lcl, le, and the power for each of the sources. 6V 2.50 W 3 Î© 1.5A 410 Î© 80 W PIT 3 Î© 30V