Question

Figure shows the net power dissipated in $R$ versus the current in a simple circuit shown.

• A. The internal resistance of battery is $0.2\Omega$.
• B. The emf of battery is $2\text{V}$
• C. $R$ at which power is $5\text{W}$ is $2.5\Omega$
• D. At $i$ = $2\text{A}$, power is $3.2\text{W}$

Answer 1

Answer: (A), (B), (D)

At $i$ = $2\text{A}$, power is $3.2\text{W}$

At max power dissipation , $r=R$

$I=\frac{E}{r+R}$
Power dissipated in $R\text{is}P=I^{2}R={\left(\frac{E}{r+R}\right)}^{2}R$
For $P$ to be maximum,
$\frac{dP}{dR}=0$
$\Rightarrow \frac{d\left[\frac{E^{2}R}{(r+R{)}^2}\right]}{dR}$=$0$
$\Rightarrow (R+r{)}^2-2R(R+r)=0$
$\Rightarrow R=r$
According to graph, $P_{max}=5\text{W}$
at $I=5\text{A}$
$\Rightarrow I^{2}R=5\text{W}$
$\Rightarrow R=r=0.2\Omega$
[At $P_{max}\Rightarrow R=r$]
Now, $\frac{E^2}{(2r{)}^2}\times r=5$
$\Rightarrow E=2\text{V}$
When , $I=2\text{A}$
Potential drop across cell is
$\Delta V_R=E-Ir$
$\Delta V_R=2\text{V}-0.4\text{V}$
$\Delta V_R=1.6\text{V}$
Power will be $I^{2}R$
=$I^2\times \frac{△V_R}{I}$
$=2\text{A}\times 1.6\text{V}=3.2\text{W}$