If E1 = 13.0 V and E2 = 5.0 V , calculate the current I2 flowing in emf source E2.
1 answer:
The solution would be like
this for this specific problem:<span>
KCL at Junction a. </span><span>
<span>+ I1 + I2 + I3 = 0 (1) </span>
<span>KVL
<span>+ 13 V - 0.2 R I1 - 0.025 R I1 - 5 V + 0.02 R I2 = 0 (2) </span>
<span>8 + 0.02 I2 = 0.225 I1 </span>
<span>I1 = 35.6 + 0.0889 I2 (2A) </span></span></span>
<span>KVL (bottom loop - CCW
direction) </span><span>
<span>- 0.02 R I2 + 5 V + 0.5 R I3 = 0 (3) </span>
<span>0.5 I3 = -5 + 0.02 I2 </span>
<span>I3 = -10 + 0.04 I2 (3A) </span></span>
<span>Replace 2A and 3A into 1. </span><span>
<span>+ I1 + I2 + I3 = 0 </span>
<span>( 35.6 + 0.0889 I2 ) + I2 + ( -10 + 0.04 I2 ) = 0 </span>
<span>1.129 I2 = -25.6 </span>
<span>I2 = -22.6A </span>
<span>Solve 2A and 3A for other currents. </span>
<span>I1 = 35.6 + 0.0889 I2 = 35.6 + 0.0889 * -22.6 = 33.5A </span>
<span>I3 = -10 + 0.04 I2 = -10 + 0.04 * -22.6 = -10.9A
So the answer is letter D.</span></span>
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Answer:
.
The unit of both sides of this equation are .
Explanation:
The unit of the left-hand side is , same as the unit of
.
The following makes use of the fact that for any non-zero value , the power
is equivalent to
.
On the right-hand side of this equation:
has no unit.
- The unit of
is
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- The unit of
is
, which is equivalent to
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- The unit of
is
.
.
Hence, the unit on the right-hand side of this equation is also .