Answer with Explanation:w
a.We are given that
Potential difference, V=48 V
[tex]R_1=24\Omega[/tex]
[tex]R_2=96\Omega[/tex]
Equivalent resistance when R1 and R2 are connected in series
[tex]R_{eq}=R_1+R_2[/tex]
Using the formula
[tex]R_{eq}=24+96=120\Omega[/tex]
We know that
[tex]I=\frac{V}{R_{eq}}=\frac{48}{120}=0.4 A[/tex]
In series combination, current passing through each resistor is same and potential difference across each resistor is different.
Power, P=[tex]I^2 R[/tex]
Using the formula
Power,[tex]P_1=I^2R_1=(0.4)^2\times 24=3.84 W[/tex]
Power, [tex]P_2=I^2 R_2=(0.4)^2(96)=15.36 W[/tex]
b.
In parallel combination, potential difference remains same across each resistor and current passing through each resistor is different..
Current,[tex]I=\frac{V}{R}[/tex]
Using the formula
[tex]I_1=\frac{V}{R_1}=\frac{48}{24}=2 A[/tex]
[tex]I_2=\frac{V}{R_2}=\frac{48}{96}=0.5 A[/tex]
[tex]P_1=\frac{V^2}{R_1}=\frac{(48)^2}{24}=96 W[/tex]
[tex]P_2=\frac{V^2}{R_2}=\frac{(48)^2}{96}=24 W[/tex]
