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3 years ago

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The voltage across the terminals of an ac power supply varies with time according to V=V0cos(t). The voltage amplitude is V0 = 4

1.0V .
A. What is the root-mean-square potential difference Vrms?
B. What is the average potential difference Vav between the two terminals of the power supply?

1 answer:

kirza4 [7]3 years ago

5 0

Answer:

A) V_rms = 29 V

B) Vav = 0 V

Explanation:

A) We are told that;

V = V_o cos ωt

voltage amplitude; V = V_o = 41.0V

Now, the formula for the root-mean-square potential difference Vrms is given as;

V_rms = V/√2

Thus plugging in relevant values, we have;

V_rms = 41/√2

V_rms = 29 V

B) Due to the fact that the voltage is sinusoidal from the given V = V_o cos ωt, we can say that the average potential difference Vav between the two terminals of the power supply would be zero.

Thus; Vav = 0 V

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Explanation:

<h2>[16]</h2>

$\underline{\boxed{\large{\bf{Option \; A!! }}}}$

Here,

$\rm { R_1}$ = 2Ω
$\rm { R_2}$ = 2Ω
$\rm { R_3}$ = 2Ω
$\rm { R_4}$ = 2Ω

We have to find the equivalent resistance of the circuit.

Here, $\rm { R_1}$ and $\rm { R_2}$ are connected in series, so their combined resistance will be given by,

$\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\$

$\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\$

$\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\$

Now, the combined resistance of $\rm { R_1}$ and $\rm { R_2}$ is connected in parallel combination with $\rm { R_3}$ , so their combined resistance will be given by,

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\$

Reciprocating both sides,

$\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\$

Now, the combined resistance of $\rm { R_1}$ , $\rm { R_2}$ and $\rm { R_3}$ is connected in series combination with $\rm { R_4}$ . So, equivalent resistance will be given by,

$\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\$

$\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\$

Henceforth, Option A is correct.

<h2>_________________________________</h2>

<h2>[17]</h2>

$\underline{\boxed{\large{\bf{Option \; B!! }}}}$

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

$\longrightarrow$ V = IR

$\longrightarrow$ 3 = I × 3.33

$\longrightarrow$ 3 ÷ 3.33 = I

$\longrightarrow$ 0.90 Ampere = I

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we apply here conservation of energy that is

1/2 × m×v² = 1/2 × k×x²

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