A current is made of two components, a DC component of I1 = 3 amp and an AC component given by I2 = 4 2 sin amp; then the reading of the hot-wire ammeter (rms value) is:

Question

A current is made of two components, a DC component of $I_{1} = 3$ amp and an AC component given by $I_{2} = 4 \sqrt{2} \sin ω t$ amp; then the reading of the hot-wire ammeter (rms value) is:

Smart Achievers · Accepted Answer

(4)

$I_{1} = 3 A,$ $I_{2} = 4 \sqrt{2} \sin ω t$

$I_{net} = I_{1} + I_{2} = (3 + 4 \sqrt{2} \sin ω t)$

$I_{rms} = \sqrt{3^{2} + {(4 \sqrt{2})}^{2} \times \frac{1}{2}} = 5 A$

Solution

Q.
A current is made of two components, a DC component of $I_{1} = 3$ amp and an AC component given by $I_{2} = 4 \sqrt{2} \sin ω t$ amp; then the reading of the hot-wire ammeter (rms value) is:

1 $3 amp$

2 $4 \sqrt{2} amp$

3 $(3 + 4 \sqrt{2}) amp$

4 $5 amp$

Ans.
(4)

$I_{1} = 3 A,$ $I_{2} = 4 \sqrt{2} \sin ω t$

$I_{net} = I_{1} + I_{2} = (3 + 4 \sqrt{2} \sin ω t)$

$I_{rms} = \sqrt{3^{2} + {(4 \sqrt{2})}^{2} \times \frac{1}{2}} = 5 A$

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Solution

Q. A current is made of two components, a DC component of I1=3 amp and an AC component given by I2=42sinωt amp; then the reading of the hot-wire ammeter (rms value) is:

1 3 amp

2 42 amp

3 (3+42) amp

4 5 amp

Ans. (4) I1=3A, I2=42sinωt Inet=I1+I2=(3+42sinωt) Irms=32+(42)2×12=5A