Question

In the circuit shown below, if $V_s = (72\angle107^\circ)$ V, find the values of (a) $I_1$ (b) $I_2$ j5 ? 3 ? -j4 ? + + + $V_s$ ~ $15\angle90^\circ$ V $20\angle20^\circ$ V (a) $I_1 = $ 15.24 ? 43.2 A (b) $I_2 = $ ? A

          In the circuit shown below, if $V_s = (72\angle107^\circ)$ V, find the values of
(a) $I_1$
(b) $I_2$

j5 ?
3 ?
-j4 ?
+
+
+
$V_s$
~
$15\angle90^\circ$ V
$20\angle20^\circ$ V
(a) $I_1 = $
15.24
? 43.2
A
(b) $I_2 = $
?
A

In the circuit shown below, if Vs = (72∠107^∘) V, find the values of
(a) I1
(b) I2

j5 ?
3 ?
-j4 ?
+
+
+
Vs

15∠90^∘ V
20∠20^∘ V
(a) I1 =
15.24
? 43.2
A
(b) I2 =
?
A

Added by Roger F.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Frank Deng

Step 1

From the circuit diagram, we can see that the voltage source V is connected in series with resistor a and resistor b. Therefore, the total voltage across the circuit is equal to the sum of the voltage drops across a and b. V = Va + Vb Since V = 72.107V, we can Show more…

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