152540
Two resistors of resistances $2 \Omega$ and $6 \Omega$ are connected in parallel. This combination is then connected to a battery of emf $2 \mathrm{~V}$ and internal resistance $0.5 \Omega$. What is the current flowing through the battery?
1 $4 \mathrm{~A}$
2 $\frac{4}{3} \mathrm{~A}$
3 $\frac{4}{17} \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
D Given, $\mathrm{R}_{1}=2 \Omega$ $\mathrm{R}_{2}=6 \Omega$ $\mathrm{r}=0.5 \Omega$ $\mathrm{E}=2 \mathrm{~V}$ Resistors $2 \Omega$ and $6 \Omega$ are connected in parallel. Their equivalent resistance is $\frac{1}{R}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$ $\mathrm{R}=\frac{3}{2} \Omega=1.5 \Omega$ Current flowing through the battery is $I=\frac{E}{R+r}$ $I=\frac{2 V}{1.5 \Omega+0.5 \Omega}=1 \mathrm{~A}$
Karnataka CET-2014
Current Electricity
152541
The current in a simple series circuit is $\mathbf{5 . 0}$ amp. When an additional resistance of $2.0 \mathrm{ohms}$ is inserted, the current drops to 4.0 amp. The original resistance of the circuit in ohms was :
1 1.25
2 8
3 10
4 20
Explanation:
B Given $I=5 \mathrm{~A}$ $\mathrm{~V}=\mathrm{I} \times \mathrm{R}$ $\mathrm{V}=5 \times \mathrm{R}$ $\mathrm{R}_{\text {total }}=\mathrm{R}+2$ $\mathrm{~V}=4(\mathrm{R}+2)$ From equation (i) and (ii) $5 R=4(R+2)$ $R=8 \Omega$ Current Electricity
Karnataka CET-2005
Current Electricity
152542
A group of $\mathbf{N}$ cells whose e.m.f. varies directly with the internal resistance as per the equation $E_{n}=1.5 r_{n}$ are connected as shown in the figure above. The current $I$ in the circuit is :
1 $5.1 \mathrm{amp}$
2 $0.51 \mathrm{amp}$
3 $1.5 \mathrm{amp}$
4 $0.15 \mathrm{amp}$
Explanation:
C Equation given as, $\mathrm{E}_{\mathrm{n}}=1.5 \mathrm{r}_{\mathrm{n}}$ We know that, $\text { Current (I) } =\frac{\sum \mathrm{E}}{\sum \mathrm{r}}$ $=\frac{1.5 \mathrm{r}_{1}+1.5 \mathrm{r}_{2}+\ldots \ldots \ldots . .5 \mathrm{r}_{\mathrm{n}}}{\mathrm{r}_{1}+\mathrm{r}_{2}+\mathrm{r}_{3}+\ldots \ldots \ldots \mathrm{r}_{\mathrm{n}}}$ $\mathrm{I} =1.5 \mathrm{~A}$
Karnataka CET-2003
Current Electricity
152543
A cell supplies a current of 0.9 A through a $2 \Omega$ resistor an a current of 0.3 A through a $7 \Omega$ resistor. The internal resistance of the cell is :
1 $1.2 \Omega$
2 $2.0 \Omega$
3 $0.5 \Omega$
4 $1.0 \Omega$
Explanation:
C Let $\mathrm{E}$ be the emf of the cell and $\mathrm{r}$ its internal resistant then, $i=\frac{E}{r+R}$ $0.9=\frac{E}{2+r}$ $0.3=\frac{E}{7+r}$ Dividing equation (i) by equation (ii) we get - $\frac{7+r}{2+r}=\frac{9}{3}$ $21+3 r=18+9 r$ $3=6 r$ $r=0.5 \Omega$ Hence the internal resistance of the cells is $0.5 \Omega$
Karnataka CET-2002
Current Electricity
152544
A battery supplies $150 \mathrm{~W}$ and $196 \mathrm{~W}$ power to two resistors of $6 \Omega$ and $4 \Omega$ when they are connected separately to it. The internal resistance of the battery is :
1 $2.5 \Omega$
2 $2 \Omega$
3 $1 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{P}_{1}=150 \mathrm{~W}, \mathrm{P}_{2}=196 \mathrm{~W}$ We know that, $\mathrm{R}_{1}=6 \Omega, \quad \mathrm{R}_{2}=4 \Omega$ $\mathrm{P} \propto \frac{\mathrm{R}}{(\mathrm{R}+\mathrm{r})^{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} \frac{\left(\mathrm{R}_{2}+\mathrm{r}\right)^{2}}{\left(\mathrm{R}_{1}+\mathrm{r}\right)^{2}}$ $\frac{25}{49}=\frac{(4+\mathrm{r})^{2}}{(6+\mathrm{r})^{2}}$ $\frac{5}{7}=\frac{4+\mathrm{r}}{6+\mathrm{r}}$ $30+5 \mathrm{r}=28+7 \mathrm{r}$ $2=2 \mathrm{r}$ $\mathrm{r}=1 \Omega$
152540
Two resistors of resistances $2 \Omega$ and $6 \Omega$ are connected in parallel. This combination is then connected to a battery of emf $2 \mathrm{~V}$ and internal resistance $0.5 \Omega$. What is the current flowing through the battery?
1 $4 \mathrm{~A}$
2 $\frac{4}{3} \mathrm{~A}$
3 $\frac{4}{17} \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
D Given, $\mathrm{R}_{1}=2 \Omega$ $\mathrm{R}_{2}=6 \Omega$ $\mathrm{r}=0.5 \Omega$ $\mathrm{E}=2 \mathrm{~V}$ Resistors $2 \Omega$ and $6 \Omega$ are connected in parallel. Their equivalent resistance is $\frac{1}{R}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$ $\mathrm{R}=\frac{3}{2} \Omega=1.5 \Omega$ Current flowing through the battery is $I=\frac{E}{R+r}$ $I=\frac{2 V}{1.5 \Omega+0.5 \Omega}=1 \mathrm{~A}$
Karnataka CET-2014
Current Electricity
152541
The current in a simple series circuit is $\mathbf{5 . 0}$ amp. When an additional resistance of $2.0 \mathrm{ohms}$ is inserted, the current drops to 4.0 amp. The original resistance of the circuit in ohms was :
1 1.25
2 8
3 10
4 20
Explanation:
B Given $I=5 \mathrm{~A}$ $\mathrm{~V}=\mathrm{I} \times \mathrm{R}$ $\mathrm{V}=5 \times \mathrm{R}$ $\mathrm{R}_{\text {total }}=\mathrm{R}+2$ $\mathrm{~V}=4(\mathrm{R}+2)$ From equation (i) and (ii) $5 R=4(R+2)$ $R=8 \Omega$ Current Electricity
Karnataka CET-2005
Current Electricity
152542
A group of $\mathbf{N}$ cells whose e.m.f. varies directly with the internal resistance as per the equation $E_{n}=1.5 r_{n}$ are connected as shown in the figure above. The current $I$ in the circuit is :
1 $5.1 \mathrm{amp}$
2 $0.51 \mathrm{amp}$
3 $1.5 \mathrm{amp}$
4 $0.15 \mathrm{amp}$
Explanation:
C Equation given as, $\mathrm{E}_{\mathrm{n}}=1.5 \mathrm{r}_{\mathrm{n}}$ We know that, $\text { Current (I) } =\frac{\sum \mathrm{E}}{\sum \mathrm{r}}$ $=\frac{1.5 \mathrm{r}_{1}+1.5 \mathrm{r}_{2}+\ldots \ldots \ldots . .5 \mathrm{r}_{\mathrm{n}}}{\mathrm{r}_{1}+\mathrm{r}_{2}+\mathrm{r}_{3}+\ldots \ldots \ldots \mathrm{r}_{\mathrm{n}}}$ $\mathrm{I} =1.5 \mathrm{~A}$
Karnataka CET-2003
Current Electricity
152543
A cell supplies a current of 0.9 A through a $2 \Omega$ resistor an a current of 0.3 A through a $7 \Omega$ resistor. The internal resistance of the cell is :
1 $1.2 \Omega$
2 $2.0 \Omega$
3 $0.5 \Omega$
4 $1.0 \Omega$
Explanation:
C Let $\mathrm{E}$ be the emf of the cell and $\mathrm{r}$ its internal resistant then, $i=\frac{E}{r+R}$ $0.9=\frac{E}{2+r}$ $0.3=\frac{E}{7+r}$ Dividing equation (i) by equation (ii) we get - $\frac{7+r}{2+r}=\frac{9}{3}$ $21+3 r=18+9 r$ $3=6 r$ $r=0.5 \Omega$ Hence the internal resistance of the cells is $0.5 \Omega$
Karnataka CET-2002
Current Electricity
152544
A battery supplies $150 \mathrm{~W}$ and $196 \mathrm{~W}$ power to two resistors of $6 \Omega$ and $4 \Omega$ when they are connected separately to it. The internal resistance of the battery is :
1 $2.5 \Omega$
2 $2 \Omega$
3 $1 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{P}_{1}=150 \mathrm{~W}, \mathrm{P}_{2}=196 \mathrm{~W}$ We know that, $\mathrm{R}_{1}=6 \Omega, \quad \mathrm{R}_{2}=4 \Omega$ $\mathrm{P} \propto \frac{\mathrm{R}}{(\mathrm{R}+\mathrm{r})^{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} \frac{\left(\mathrm{R}_{2}+\mathrm{r}\right)^{2}}{\left(\mathrm{R}_{1}+\mathrm{r}\right)^{2}}$ $\frac{25}{49}=\frac{(4+\mathrm{r})^{2}}{(6+\mathrm{r})^{2}}$ $\frac{5}{7}=\frac{4+\mathrm{r}}{6+\mathrm{r}}$ $30+5 \mathrm{r}=28+7 \mathrm{r}$ $2=2 \mathrm{r}$ $\mathrm{r}=1 \Omega$
152540
Two resistors of resistances $2 \Omega$ and $6 \Omega$ are connected in parallel. This combination is then connected to a battery of emf $2 \mathrm{~V}$ and internal resistance $0.5 \Omega$. What is the current flowing through the battery?
1 $4 \mathrm{~A}$
2 $\frac{4}{3} \mathrm{~A}$
3 $\frac{4}{17} \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
D Given, $\mathrm{R}_{1}=2 \Omega$ $\mathrm{R}_{2}=6 \Omega$ $\mathrm{r}=0.5 \Omega$ $\mathrm{E}=2 \mathrm{~V}$ Resistors $2 \Omega$ and $6 \Omega$ are connected in parallel. Their equivalent resistance is $\frac{1}{R}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$ $\mathrm{R}=\frac{3}{2} \Omega=1.5 \Omega$ Current flowing through the battery is $I=\frac{E}{R+r}$ $I=\frac{2 V}{1.5 \Omega+0.5 \Omega}=1 \mathrm{~A}$
Karnataka CET-2014
Current Electricity
152541
The current in a simple series circuit is $\mathbf{5 . 0}$ amp. When an additional resistance of $2.0 \mathrm{ohms}$ is inserted, the current drops to 4.0 amp. The original resistance of the circuit in ohms was :
1 1.25
2 8
3 10
4 20
Explanation:
B Given $I=5 \mathrm{~A}$ $\mathrm{~V}=\mathrm{I} \times \mathrm{R}$ $\mathrm{V}=5 \times \mathrm{R}$ $\mathrm{R}_{\text {total }}=\mathrm{R}+2$ $\mathrm{~V}=4(\mathrm{R}+2)$ From equation (i) and (ii) $5 R=4(R+2)$ $R=8 \Omega$ Current Electricity
Karnataka CET-2005
Current Electricity
152542
A group of $\mathbf{N}$ cells whose e.m.f. varies directly with the internal resistance as per the equation $E_{n}=1.5 r_{n}$ are connected as shown in the figure above. The current $I$ in the circuit is :
1 $5.1 \mathrm{amp}$
2 $0.51 \mathrm{amp}$
3 $1.5 \mathrm{amp}$
4 $0.15 \mathrm{amp}$
Explanation:
C Equation given as, $\mathrm{E}_{\mathrm{n}}=1.5 \mathrm{r}_{\mathrm{n}}$ We know that, $\text { Current (I) } =\frac{\sum \mathrm{E}}{\sum \mathrm{r}}$ $=\frac{1.5 \mathrm{r}_{1}+1.5 \mathrm{r}_{2}+\ldots \ldots \ldots . .5 \mathrm{r}_{\mathrm{n}}}{\mathrm{r}_{1}+\mathrm{r}_{2}+\mathrm{r}_{3}+\ldots \ldots \ldots \mathrm{r}_{\mathrm{n}}}$ $\mathrm{I} =1.5 \mathrm{~A}$
Karnataka CET-2003
Current Electricity
152543
A cell supplies a current of 0.9 A through a $2 \Omega$ resistor an a current of 0.3 A through a $7 \Omega$ resistor. The internal resistance of the cell is :
1 $1.2 \Omega$
2 $2.0 \Omega$
3 $0.5 \Omega$
4 $1.0 \Omega$
Explanation:
C Let $\mathrm{E}$ be the emf of the cell and $\mathrm{r}$ its internal resistant then, $i=\frac{E}{r+R}$ $0.9=\frac{E}{2+r}$ $0.3=\frac{E}{7+r}$ Dividing equation (i) by equation (ii) we get - $\frac{7+r}{2+r}=\frac{9}{3}$ $21+3 r=18+9 r$ $3=6 r$ $r=0.5 \Omega$ Hence the internal resistance of the cells is $0.5 \Omega$
Karnataka CET-2002
Current Electricity
152544
A battery supplies $150 \mathrm{~W}$ and $196 \mathrm{~W}$ power to two resistors of $6 \Omega$ and $4 \Omega$ when they are connected separately to it. The internal resistance of the battery is :
1 $2.5 \Omega$
2 $2 \Omega$
3 $1 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{P}_{1}=150 \mathrm{~W}, \mathrm{P}_{2}=196 \mathrm{~W}$ We know that, $\mathrm{R}_{1}=6 \Omega, \quad \mathrm{R}_{2}=4 \Omega$ $\mathrm{P} \propto \frac{\mathrm{R}}{(\mathrm{R}+\mathrm{r})^{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} \frac{\left(\mathrm{R}_{2}+\mathrm{r}\right)^{2}}{\left(\mathrm{R}_{1}+\mathrm{r}\right)^{2}}$ $\frac{25}{49}=\frac{(4+\mathrm{r})^{2}}{(6+\mathrm{r})^{2}}$ $\frac{5}{7}=\frac{4+\mathrm{r}}{6+\mathrm{r}}$ $30+5 \mathrm{r}=28+7 \mathrm{r}$ $2=2 \mathrm{r}$ $\mathrm{r}=1 \Omega$
152540
Two resistors of resistances $2 \Omega$ and $6 \Omega$ are connected in parallel. This combination is then connected to a battery of emf $2 \mathrm{~V}$ and internal resistance $0.5 \Omega$. What is the current flowing through the battery?
1 $4 \mathrm{~A}$
2 $\frac{4}{3} \mathrm{~A}$
3 $\frac{4}{17} \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
D Given, $\mathrm{R}_{1}=2 \Omega$ $\mathrm{R}_{2}=6 \Omega$ $\mathrm{r}=0.5 \Omega$ $\mathrm{E}=2 \mathrm{~V}$ Resistors $2 \Omega$ and $6 \Omega$ are connected in parallel. Their equivalent resistance is $\frac{1}{R}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$ $\mathrm{R}=\frac{3}{2} \Omega=1.5 \Omega$ Current flowing through the battery is $I=\frac{E}{R+r}$ $I=\frac{2 V}{1.5 \Omega+0.5 \Omega}=1 \mathrm{~A}$
Karnataka CET-2014
Current Electricity
152541
The current in a simple series circuit is $\mathbf{5 . 0}$ amp. When an additional resistance of $2.0 \mathrm{ohms}$ is inserted, the current drops to 4.0 amp. The original resistance of the circuit in ohms was :
1 1.25
2 8
3 10
4 20
Explanation:
B Given $I=5 \mathrm{~A}$ $\mathrm{~V}=\mathrm{I} \times \mathrm{R}$ $\mathrm{V}=5 \times \mathrm{R}$ $\mathrm{R}_{\text {total }}=\mathrm{R}+2$ $\mathrm{~V}=4(\mathrm{R}+2)$ From equation (i) and (ii) $5 R=4(R+2)$ $R=8 \Omega$ Current Electricity
Karnataka CET-2005
Current Electricity
152542
A group of $\mathbf{N}$ cells whose e.m.f. varies directly with the internal resistance as per the equation $E_{n}=1.5 r_{n}$ are connected as shown in the figure above. The current $I$ in the circuit is :
1 $5.1 \mathrm{amp}$
2 $0.51 \mathrm{amp}$
3 $1.5 \mathrm{amp}$
4 $0.15 \mathrm{amp}$
Explanation:
C Equation given as, $\mathrm{E}_{\mathrm{n}}=1.5 \mathrm{r}_{\mathrm{n}}$ We know that, $\text { Current (I) } =\frac{\sum \mathrm{E}}{\sum \mathrm{r}}$ $=\frac{1.5 \mathrm{r}_{1}+1.5 \mathrm{r}_{2}+\ldots \ldots \ldots . .5 \mathrm{r}_{\mathrm{n}}}{\mathrm{r}_{1}+\mathrm{r}_{2}+\mathrm{r}_{3}+\ldots \ldots \ldots \mathrm{r}_{\mathrm{n}}}$ $\mathrm{I} =1.5 \mathrm{~A}$
Karnataka CET-2003
Current Electricity
152543
A cell supplies a current of 0.9 A through a $2 \Omega$ resistor an a current of 0.3 A through a $7 \Omega$ resistor. The internal resistance of the cell is :
1 $1.2 \Omega$
2 $2.0 \Omega$
3 $0.5 \Omega$
4 $1.0 \Omega$
Explanation:
C Let $\mathrm{E}$ be the emf of the cell and $\mathrm{r}$ its internal resistant then, $i=\frac{E}{r+R}$ $0.9=\frac{E}{2+r}$ $0.3=\frac{E}{7+r}$ Dividing equation (i) by equation (ii) we get - $\frac{7+r}{2+r}=\frac{9}{3}$ $21+3 r=18+9 r$ $3=6 r$ $r=0.5 \Omega$ Hence the internal resistance of the cells is $0.5 \Omega$
Karnataka CET-2002
Current Electricity
152544
A battery supplies $150 \mathrm{~W}$ and $196 \mathrm{~W}$ power to two resistors of $6 \Omega$ and $4 \Omega$ when they are connected separately to it. The internal resistance of the battery is :
1 $2.5 \Omega$
2 $2 \Omega$
3 $1 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{P}_{1}=150 \mathrm{~W}, \mathrm{P}_{2}=196 \mathrm{~W}$ We know that, $\mathrm{R}_{1}=6 \Omega, \quad \mathrm{R}_{2}=4 \Omega$ $\mathrm{P} \propto \frac{\mathrm{R}}{(\mathrm{R}+\mathrm{r})^{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} \frac{\left(\mathrm{R}_{2}+\mathrm{r}\right)^{2}}{\left(\mathrm{R}_{1}+\mathrm{r}\right)^{2}}$ $\frac{25}{49}=\frac{(4+\mathrm{r})^{2}}{(6+\mathrm{r})^{2}}$ $\frac{5}{7}=\frac{4+\mathrm{r}}{6+\mathrm{r}}$ $30+5 \mathrm{r}=28+7 \mathrm{r}$ $2=2 \mathrm{r}$ $\mathrm{r}=1 \Omega$
152540
Two resistors of resistances $2 \Omega$ and $6 \Omega$ are connected in parallel. This combination is then connected to a battery of emf $2 \mathrm{~V}$ and internal resistance $0.5 \Omega$. What is the current flowing through the battery?
1 $4 \mathrm{~A}$
2 $\frac{4}{3} \mathrm{~A}$
3 $\frac{4}{17} \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
D Given, $\mathrm{R}_{1}=2 \Omega$ $\mathrm{R}_{2}=6 \Omega$ $\mathrm{r}=0.5 \Omega$ $\mathrm{E}=2 \mathrm{~V}$ Resistors $2 \Omega$ and $6 \Omega$ are connected in parallel. Their equivalent resistance is $\frac{1}{R}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$ $\mathrm{R}=\frac{3}{2} \Omega=1.5 \Omega$ Current flowing through the battery is $I=\frac{E}{R+r}$ $I=\frac{2 V}{1.5 \Omega+0.5 \Omega}=1 \mathrm{~A}$
Karnataka CET-2014
Current Electricity
152541
The current in a simple series circuit is $\mathbf{5 . 0}$ amp. When an additional resistance of $2.0 \mathrm{ohms}$ is inserted, the current drops to 4.0 amp. The original resistance of the circuit in ohms was :
1 1.25
2 8
3 10
4 20
Explanation:
B Given $I=5 \mathrm{~A}$ $\mathrm{~V}=\mathrm{I} \times \mathrm{R}$ $\mathrm{V}=5 \times \mathrm{R}$ $\mathrm{R}_{\text {total }}=\mathrm{R}+2$ $\mathrm{~V}=4(\mathrm{R}+2)$ From equation (i) and (ii) $5 R=4(R+2)$ $R=8 \Omega$ Current Electricity
Karnataka CET-2005
Current Electricity
152542
A group of $\mathbf{N}$ cells whose e.m.f. varies directly with the internal resistance as per the equation $E_{n}=1.5 r_{n}$ are connected as shown in the figure above. The current $I$ in the circuit is :
1 $5.1 \mathrm{amp}$
2 $0.51 \mathrm{amp}$
3 $1.5 \mathrm{amp}$
4 $0.15 \mathrm{amp}$
Explanation:
C Equation given as, $\mathrm{E}_{\mathrm{n}}=1.5 \mathrm{r}_{\mathrm{n}}$ We know that, $\text { Current (I) } =\frac{\sum \mathrm{E}}{\sum \mathrm{r}}$ $=\frac{1.5 \mathrm{r}_{1}+1.5 \mathrm{r}_{2}+\ldots \ldots \ldots . .5 \mathrm{r}_{\mathrm{n}}}{\mathrm{r}_{1}+\mathrm{r}_{2}+\mathrm{r}_{3}+\ldots \ldots \ldots \mathrm{r}_{\mathrm{n}}}$ $\mathrm{I} =1.5 \mathrm{~A}$
Karnataka CET-2003
Current Electricity
152543
A cell supplies a current of 0.9 A through a $2 \Omega$ resistor an a current of 0.3 A through a $7 \Omega$ resistor. The internal resistance of the cell is :
1 $1.2 \Omega$
2 $2.0 \Omega$
3 $0.5 \Omega$
4 $1.0 \Omega$
Explanation:
C Let $\mathrm{E}$ be the emf of the cell and $\mathrm{r}$ its internal resistant then, $i=\frac{E}{r+R}$ $0.9=\frac{E}{2+r}$ $0.3=\frac{E}{7+r}$ Dividing equation (i) by equation (ii) we get - $\frac{7+r}{2+r}=\frac{9}{3}$ $21+3 r=18+9 r$ $3=6 r$ $r=0.5 \Omega$ Hence the internal resistance of the cells is $0.5 \Omega$
Karnataka CET-2002
Current Electricity
152544
A battery supplies $150 \mathrm{~W}$ and $196 \mathrm{~W}$ power to two resistors of $6 \Omega$ and $4 \Omega$ when they are connected separately to it. The internal resistance of the battery is :
1 $2.5 \Omega$
2 $2 \Omega$
3 $1 \Omega$
4 $0.5 \Omega$
Explanation:
C Given that, $\mathrm{P}_{1}=150 \mathrm{~W}, \mathrm{P}_{2}=196 \mathrm{~W}$ We know that, $\mathrm{R}_{1}=6 \Omega, \quad \mathrm{R}_{2}=4 \Omega$ $\mathrm{P} \propto \frac{\mathrm{R}}{(\mathrm{R}+\mathrm{r})^{2}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}} \frac{\left(\mathrm{R}_{2}+\mathrm{r}\right)^{2}}{\left(\mathrm{R}_{1}+\mathrm{r}\right)^{2}}$ $\frac{25}{49}=\frac{(4+\mathrm{r})^{2}}{(6+\mathrm{r})^{2}}$ $\frac{5}{7}=\frac{4+\mathrm{r}}{6+\mathrm{r}}$ $30+5 \mathrm{r}=28+7 \mathrm{r}$ $2=2 \mathrm{r}$ $\mathrm{r}=1 \Omega$
