Physics, 03.03.2020 04:49 heyitzmeamelie

(a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.

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Answer from: biaxialpower789

Answer with Explanation:w

a.We are given that

Potential difference, V=48 V

$R_1=24\Omega$

$R_2=96\Omega$

Equivalent resistance when R1 and R2 are connected in series

$R_{eq}=R_1+R_2$

Using the formula

$R_{eq}=24+96=120\Omega$

We know that

$I=\frac{V}{R_{eq}}=\frac{48}{120}=0.4 A$

In series combination, current passing through each resistor is same and potential difference across each resistor is different.

Power, P= I^2 R

Using the formula

Power, $P_1=I^2R_1=(0.4)^2\times 24=3.84 W$

Power, P_2=I^2 R_2=(0.4)^2(96)=15.36 W

In parallel combination, potential difference remains same across each resistor and current passing through each resistor is different..

Current, $I=\frac{V}{R}$

Using the formula

$I_1=\frac{V}{R_1}=\frac{48}{24}=2 A$

$I_2=\frac{V}{R_2}=\frac{48}{96}=0.5 A$

$P_1=\frac{V^2}{R_1}=\frac{(48)^2}{24}=96 W$

$P_2=\frac{V^2}{R_2}=\frac{(48)^2}{96}=24 W$

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(a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when...

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