Two cells of same emf $E$ but internal resistance $r_1$ and $r_2$ are connected in series to an external resistor $R$ (figure). What should be the value of $R$ so that the potential difference across the terminals of the first cell becomes zero?
Applying Ohm's law,
Effective resistance $=R+r_1+r_2$ and effective emf of two cells $=E+E=2 E$, so the electric current is given by
$$I=\frac{E+E}{R+r_1+r_2}$$
The potential difference across the terminals of the first cell and putting it equal to zero.
$$\begin{aligned} & V=E-I_1=E-\frac{2 E}{r_1+r_2+R} r_1=0 \\ \text{or}\quad & E=\frac{2 E r_1}{r_1+r_2+R} \Rightarrow 1=\frac{2 r_1}{r_1+r_2+R} \\ & r_1+r_2+R=2 r_1 \Rightarrow R=r_1-r_2 \end{aligned}$$
This is the required relation.
Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter 1 mm . Conductor $B$ is a hollow tube of outer diameter 2 mm and inner diameter 1 mm . Find the ratio of resistance $R_A$ to $R_B$.
The resistance of first conductor
$$R_A=\frac{\rho l}{\pi\left(10^{-3} \times 0.5\right)^2}$$
The resistance of second conductor,
$$R_B=\frac{\rho l}{\pi\left[\left(10^{-3}\right)^2-\left(0.5 \times 10^{-3}\right)^2\right]}$$
Now, the ratio of two resistors is given by
$$\frac{R_A}{R_B}=\frac{\left(10^{-3}\right)^2-\left(0.5 \times 10^{-3}\right)^2}{\left(0.5 \times 10^{-3}\right)^2}=3: 1$$
Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to $n$-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Examples 3,7 in the NCERT Text Book for Class XII.
Let the effective internal resistance of the battery is $R_{\text {eff }}$, the effective external resistance $R$ and the effective voltage of the battery is $V_{\text {eff }}$.
Applying Ohm's law,
Then current through $R$ is given by
$$I=\frac{V_{\text {eff }}}{R_{\text {eff }}+R}$$
If all the resistances and the effective voltage are increased $n$-times, then we have
$$\begin{aligned} V_{\text {eff }}^{\text {new }} & =n V_{\text {eff }}, R_{\text {eff }}^{\text {new }}=n R_{\text {eff }} \\ \text{and}\quad R^{\text {new }} & =n R \end{aligned}$$
Then, the new current is given by
$$I^{\prime}=\frac{n V_{\text {eff }}}{n R_{\text {eff }}+n R}=\frac{n\left(V_{\text {eff }}\right)}{n\left(R_{\text {eff }}+R\right)}=\frac{\left(V_{\text {eff }}\right)}{\left(R_{\text {eff }}+R\right)}=I$$
Thus, current remains the same.
Two cells of voltage 10 V and 2 V and $10 \Omega$ internal resistances $10 \Omega$ and $5 \Omega$ respectively, are connected in parallel with the positive end of 10 V battery connected to negative pole of 2 V battery (figure). Find the effective voltage and effective resistance of the combination.
Applying Kirchhoff's junction rule, $\quad I_1=I+I_2$
Applying Kirchhoff's II law / loop rule applied in outer loop containing 10V cell and resistance $R$, we have
$$10=I R+10 I_1\quad\text{.... (i)}$$
Applying Kirchhoff II law / loop rule applied in outer loop containing 2 V cell and resistance $R$, we have
$$\begin{aligned} 2 & =5 I_2-R I=5\left(I_1-I\right)-R I \\ \text{or}\quad 4 & =10 I_1-10 I-2 R I\quad \text{... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { Solving Eqs. (i) and (ii), gives }\\ &\begin{aligned} & \Rightarrow \quad 6=3 R I+10 I \\ & 2=I\left(R+\frac{10}{3}\right) \end{aligned} \end{aligned}$$
Also, the external resistance is $R$. The Ohm's law states that
$$V=I\left(R+R_{\text {eff }}\right)$$
On comparing, we have $V=2 V$ and effective internal resistance
$$\left(R_{\text {eff }}\right)=\left(\frac{10}{3}\right) \Omega$$
Since, the effective internal resistance $\left(R_{\text {eff }}\right)$ of two cells is $\left(\frac{10}{3}\right) \Omega$, being the parallel combination of $5 \Omega$ and $10 \Omega$. The equivalent circuit is given below
A room has $A C$ run for 5 a day at a voltage of 220 V . The wiring of the room consists of Cu of 1 mm radius and a length of 10 m . Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?
$$\left[\rho_{\mathrm{Cu}}=11.7 \times 10^{-8} \Omega \mathrm{~m}, \rho_{\mathrm{Al}}=2.7 \times 10^{-8} \Omega \mathrm{~m}\right]$$
Power consumption in a day i.e., in $5=10$ units
Or power consumption per hour $=2$ units
Or power consumption $=2$ units $=2 \mathrm{~kW}=2000 \mathrm{~J} / \mathrm{s}$
Also, we know that power consumption in resistor,
$$\begin{array}{rlrl} P =V \times l \\ \Rightarrow \quad 2000 \mathrm{~W} & =220 \mathrm{~V} \times l \text { or } l \approx 9 \mathrm{~A} \end{array}$$
Now, the resistance of wire is given by $R=\rho \frac{l}{A}$ where, $A$ is cross-sectional area of conductor.
Power consumption in first current carrying wire is given by
$$\begin{aligned} P & =l^2 R \\ \rho \frac{l}{A} l^2 & =1.7 \times 10^{-8} \times \frac{10}{\pi \times 10^{-6}} \times 81 \mathrm{~J} / \mathrm{s} \approx 4 \mathrm{~J} / \mathrm{s} \end{aligned}$$
The fractional loss due to the joule heating in first wire $=\frac{4}{2000} \times 100=0.2 \%$
Power loss in Al wire $=4 \frac{\rho_{\mathrm{Al}}}{\rho_{\mathrm{Cu}}}=1.6 \times 4=6.4 \mathrm{~J} / \mathrm{s}$
The fractional loss due to the joule heating in second wire $=\frac{6.4}{2000} \times 100=0.32 \%$