Question

Three resistors having resistance of $1.6\Omega ,2.4\Omega ,$ and $4.8\Omega$ are connected in series to a $22V$ battery. Find

1. The current in each resistor.
2. The potential difference across each resistor.
3. The total current through the battery.
4. The power dissipated by each resistor.

Which resistor dissipates maximum power?

Answer 1

Given data:

Three Resistance are $R_1=1.6\Omega$ , $R_2=2.4\Omega$ ,$R_3=4.8\Omega$

Voltage is$V=22V$

a) The current in each resistor:

$I=\frac{V}{R_1+R_2+R_3}$

$$\begin{gathered} =\frac{22}{1.6+2.4+4.8} \\ =\frac{22}{8.8} \end{gathered}$$

$=2.5A$

b) The potential difference across each resistor:

$V_1=I{R}_1$

$$\begin{gathered} =2.5\times 1.6 \\ =4\nu \end{gathered}$$

$V_2=I{R}_2$

$$\begin{gathered} =2.5\times 2.4 \\ =6\nu \end{gathered}$$

$V_3=I{R}_3$

$$\begin{gathered} =2.5\times 4.8 \\ 12\nu \end{gathered}$$

C) The total current through the battery:

$I=\frac{V}{R_1+R_2+R_3}$

$$\begin{gathered} =\frac{22}{1.6+2.4+4.8} \\ =\frac{22}{8.8} \end{gathered}$$

$=2.5A$

d)The power dissipated by each resistor:

$P_1=I^2{R}_1$

$$\begin{gathered} ={(2.5)}^2(1.6) \\ =10W \end{gathered}$$

$P_2=I^2{R}_2$

$$\begin{gathered} ={(2.5)}^2(2.4) \\ =15W \end{gathered}$$

$P_3=I^2{R}_3$

$$\begin{gathered} ={(2.5)}^2(4.8) \\ 30W \end{gathered}$$

e) Which resistor dissipates maximum power?
In a series connection, the maximum power is consumed by the greatest power.

$R=4.8\Omega$

Final answer:
a) Current in each resistor is $2.5A$.

b) Potential difference across each resistor is ${\nu }_1=4v,v_2=6v,v_3=12v$

c) Total current through the battery is $2.5A$.

d) Power dissipated by each resistor is $p_1=10w,p_2=15w,p_3=30w$

e)The resistor dissipates maximum power is $4.8\Omega$