151998
A $30 \mathrm{~V}, 90 \mathrm{~W}$ lamp is to be operated on a $120 \mathrm{~V}$ D.C. line. For proper glow, a resistor of ........ ohm should be connected in series with the lamp.
1 40
2 10
3 20
4 30
Explanation:
D Given, Resistance of the lamp $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{30^{2}}{90}=10 \Omega$ Current flowing the lamp $(\mathrm{I})=\frac{\mathrm{P}}{\mathrm{V}}=\frac{90}{30}=3 \mathrm{~A}$ Resistance of the line, $\mathrm{R}_{\mathrm{d}}=\frac{\mathrm{V}_{\mathrm{d}}}{\mathrm{I}}=\frac{120}{3}=40 \Omega$ If $\mathrm{R}^{\prime}$ is the resistor added, $R_{d}=R^{\prime}+R$ $R^{\prime}=R_{d}-R=40-10=30 \Omega$ Hence, for proper glow of lamp a $30 \Omega$ resistor.
Karnataka CET-2006
Current Electricity
151999
In the given circuit, the voltmeter records 5 volts. The resistance of the voltmeter in ohms is:
1 200
2 100
3 10
4 50
Explanation:
B Total potential of $10 \mathrm{~V}$ equally distributes between $50 \Omega$ and other parallel combination of $100 \Omega$ and voltmeter. Resistance of voltmeter $=\frac{100 \times \mathrm{R}}{100+\mathrm{R}}=50$ $\mathrm{R}=100 \Omega$
Karnataka CET-2005
Current Electricity
152000
Consider a rectangular slab of length $L$ and area of cross section $A$. A current $I$ is passed through it. If the length is doubled, the potential drop across the end faces
1 becomes half of the initial value
2 becomes one-fourth of the initial value
3 becomes double the initial value
4 remains same
Explanation:
C Resistance, $R=\frac{\rho l}{A}$ $\mathrm{V}=\mathrm{IR}$ $\mathrm{V}=\mathrm{I} \times \frac{\rho l}{\mathrm{~A}}$ Now, when length is doubled, $\mathrm{V}^{\prime} =\mathrm{I} \times \rho \times \frac{2 l}{\mathrm{~A}}$ $\mathrm{~V}^{\prime} =2 \mathrm{~V}$ Become double the initial value
Current Electricity
152002
A given piece of wire of length $\mathrm{l}$, cross-sectional area $A$ has a resistance $R$. The wire is stretched uniformly to a new length 21 . What is the resistance of the wire now?
152005
Two filaments of same length are connected first in series and then in parallel. For the same amount of main current flowing the ratio of the heat produced is
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
C When filament connected in series, $R_{\text {eq }}=R_{1}+R_{2}$ $R_{e q}=\frac{\rho L}{A}+\frac{\rho L}{A}=\frac{2 \rho L}{A}$ Power $\left(\mathrm{P}_{1}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}=\mathrm{I}^{2} \frac{2 \rho \mathrm{L}}{\mathrm{A}}$ When filament connected in parallel, $\mathrm{R}_{\text {eq }}^{\prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\frac{\mathrm{R} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}^{2}}{2 \mathrm{R}}=\frac{\mathrm{R}}{2}=\frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ Power $\left(\mathrm{P}_{2}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}^{\prime}=\mathrm{I}^{2} \frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}} =\frac{\mathrm{I}^{2} 2 \rho \mathrm{L}}{\mathrm{A}} \times \frac{2 \mathrm{~A}}{\rho \mathrm{LI}^{2}}$ $=4: 1$
151998
A $30 \mathrm{~V}, 90 \mathrm{~W}$ lamp is to be operated on a $120 \mathrm{~V}$ D.C. line. For proper glow, a resistor of ........ ohm should be connected in series with the lamp.
1 40
2 10
3 20
4 30
Explanation:
D Given, Resistance of the lamp $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{30^{2}}{90}=10 \Omega$ Current flowing the lamp $(\mathrm{I})=\frac{\mathrm{P}}{\mathrm{V}}=\frac{90}{30}=3 \mathrm{~A}$ Resistance of the line, $\mathrm{R}_{\mathrm{d}}=\frac{\mathrm{V}_{\mathrm{d}}}{\mathrm{I}}=\frac{120}{3}=40 \Omega$ If $\mathrm{R}^{\prime}$ is the resistor added, $R_{d}=R^{\prime}+R$ $R^{\prime}=R_{d}-R=40-10=30 \Omega$ Hence, for proper glow of lamp a $30 \Omega$ resistor.
Karnataka CET-2006
Current Electricity
151999
In the given circuit, the voltmeter records 5 volts. The resistance of the voltmeter in ohms is:
1 200
2 100
3 10
4 50
Explanation:
B Total potential of $10 \mathrm{~V}$ equally distributes between $50 \Omega$ and other parallel combination of $100 \Omega$ and voltmeter. Resistance of voltmeter $=\frac{100 \times \mathrm{R}}{100+\mathrm{R}}=50$ $\mathrm{R}=100 \Omega$
Karnataka CET-2005
Current Electricity
152000
Consider a rectangular slab of length $L$ and area of cross section $A$. A current $I$ is passed through it. If the length is doubled, the potential drop across the end faces
1 becomes half of the initial value
2 becomes one-fourth of the initial value
3 becomes double the initial value
4 remains same
Explanation:
C Resistance, $R=\frac{\rho l}{A}$ $\mathrm{V}=\mathrm{IR}$ $\mathrm{V}=\mathrm{I} \times \frac{\rho l}{\mathrm{~A}}$ Now, when length is doubled, $\mathrm{V}^{\prime} =\mathrm{I} \times \rho \times \frac{2 l}{\mathrm{~A}}$ $\mathrm{~V}^{\prime} =2 \mathrm{~V}$ Become double the initial value
Current Electricity
152002
A given piece of wire of length $\mathrm{l}$, cross-sectional area $A$ has a resistance $R$. The wire is stretched uniformly to a new length 21 . What is the resistance of the wire now?
152005
Two filaments of same length are connected first in series and then in parallel. For the same amount of main current flowing the ratio of the heat produced is
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
C When filament connected in series, $R_{\text {eq }}=R_{1}+R_{2}$ $R_{e q}=\frac{\rho L}{A}+\frac{\rho L}{A}=\frac{2 \rho L}{A}$ Power $\left(\mathrm{P}_{1}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}=\mathrm{I}^{2} \frac{2 \rho \mathrm{L}}{\mathrm{A}}$ When filament connected in parallel, $\mathrm{R}_{\text {eq }}^{\prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\frac{\mathrm{R} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}^{2}}{2 \mathrm{R}}=\frac{\mathrm{R}}{2}=\frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ Power $\left(\mathrm{P}_{2}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}^{\prime}=\mathrm{I}^{2} \frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}} =\frac{\mathrm{I}^{2} 2 \rho \mathrm{L}}{\mathrm{A}} \times \frac{2 \mathrm{~A}}{\rho \mathrm{LI}^{2}}$ $=4: 1$
151998
A $30 \mathrm{~V}, 90 \mathrm{~W}$ lamp is to be operated on a $120 \mathrm{~V}$ D.C. line. For proper glow, a resistor of ........ ohm should be connected in series with the lamp.
1 40
2 10
3 20
4 30
Explanation:
D Given, Resistance of the lamp $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{30^{2}}{90}=10 \Omega$ Current flowing the lamp $(\mathrm{I})=\frac{\mathrm{P}}{\mathrm{V}}=\frac{90}{30}=3 \mathrm{~A}$ Resistance of the line, $\mathrm{R}_{\mathrm{d}}=\frac{\mathrm{V}_{\mathrm{d}}}{\mathrm{I}}=\frac{120}{3}=40 \Omega$ If $\mathrm{R}^{\prime}$ is the resistor added, $R_{d}=R^{\prime}+R$ $R^{\prime}=R_{d}-R=40-10=30 \Omega$ Hence, for proper glow of lamp a $30 \Omega$ resistor.
Karnataka CET-2006
Current Electricity
151999
In the given circuit, the voltmeter records 5 volts. The resistance of the voltmeter in ohms is:
1 200
2 100
3 10
4 50
Explanation:
B Total potential of $10 \mathrm{~V}$ equally distributes between $50 \Omega$ and other parallel combination of $100 \Omega$ and voltmeter. Resistance of voltmeter $=\frac{100 \times \mathrm{R}}{100+\mathrm{R}}=50$ $\mathrm{R}=100 \Omega$
Karnataka CET-2005
Current Electricity
152000
Consider a rectangular slab of length $L$ and area of cross section $A$. A current $I$ is passed through it. If the length is doubled, the potential drop across the end faces
1 becomes half of the initial value
2 becomes one-fourth of the initial value
3 becomes double the initial value
4 remains same
Explanation:
C Resistance, $R=\frac{\rho l}{A}$ $\mathrm{V}=\mathrm{IR}$ $\mathrm{V}=\mathrm{I} \times \frac{\rho l}{\mathrm{~A}}$ Now, when length is doubled, $\mathrm{V}^{\prime} =\mathrm{I} \times \rho \times \frac{2 l}{\mathrm{~A}}$ $\mathrm{~V}^{\prime} =2 \mathrm{~V}$ Become double the initial value
Current Electricity
152002
A given piece of wire of length $\mathrm{l}$, cross-sectional area $A$ has a resistance $R$. The wire is stretched uniformly to a new length 21 . What is the resistance of the wire now?
152005
Two filaments of same length are connected first in series and then in parallel. For the same amount of main current flowing the ratio of the heat produced is
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
C When filament connected in series, $R_{\text {eq }}=R_{1}+R_{2}$ $R_{e q}=\frac{\rho L}{A}+\frac{\rho L}{A}=\frac{2 \rho L}{A}$ Power $\left(\mathrm{P}_{1}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}=\mathrm{I}^{2} \frac{2 \rho \mathrm{L}}{\mathrm{A}}$ When filament connected in parallel, $\mathrm{R}_{\text {eq }}^{\prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\frac{\mathrm{R} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}^{2}}{2 \mathrm{R}}=\frac{\mathrm{R}}{2}=\frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ Power $\left(\mathrm{P}_{2}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}^{\prime}=\mathrm{I}^{2} \frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}} =\frac{\mathrm{I}^{2} 2 \rho \mathrm{L}}{\mathrm{A}} \times \frac{2 \mathrm{~A}}{\rho \mathrm{LI}^{2}}$ $=4: 1$
151998
A $30 \mathrm{~V}, 90 \mathrm{~W}$ lamp is to be operated on a $120 \mathrm{~V}$ D.C. line. For proper glow, a resistor of ........ ohm should be connected in series with the lamp.
1 40
2 10
3 20
4 30
Explanation:
D Given, Resistance of the lamp $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{30^{2}}{90}=10 \Omega$ Current flowing the lamp $(\mathrm{I})=\frac{\mathrm{P}}{\mathrm{V}}=\frac{90}{30}=3 \mathrm{~A}$ Resistance of the line, $\mathrm{R}_{\mathrm{d}}=\frac{\mathrm{V}_{\mathrm{d}}}{\mathrm{I}}=\frac{120}{3}=40 \Omega$ If $\mathrm{R}^{\prime}$ is the resistor added, $R_{d}=R^{\prime}+R$ $R^{\prime}=R_{d}-R=40-10=30 \Omega$ Hence, for proper glow of lamp a $30 \Omega$ resistor.
Karnataka CET-2006
Current Electricity
151999
In the given circuit, the voltmeter records 5 volts. The resistance of the voltmeter in ohms is:
1 200
2 100
3 10
4 50
Explanation:
B Total potential of $10 \mathrm{~V}$ equally distributes between $50 \Omega$ and other parallel combination of $100 \Omega$ and voltmeter. Resistance of voltmeter $=\frac{100 \times \mathrm{R}}{100+\mathrm{R}}=50$ $\mathrm{R}=100 \Omega$
Karnataka CET-2005
Current Electricity
152000
Consider a rectangular slab of length $L$ and area of cross section $A$. A current $I$ is passed through it. If the length is doubled, the potential drop across the end faces
1 becomes half of the initial value
2 becomes one-fourth of the initial value
3 becomes double the initial value
4 remains same
Explanation:
C Resistance, $R=\frac{\rho l}{A}$ $\mathrm{V}=\mathrm{IR}$ $\mathrm{V}=\mathrm{I} \times \frac{\rho l}{\mathrm{~A}}$ Now, when length is doubled, $\mathrm{V}^{\prime} =\mathrm{I} \times \rho \times \frac{2 l}{\mathrm{~A}}$ $\mathrm{~V}^{\prime} =2 \mathrm{~V}$ Become double the initial value
Current Electricity
152002
A given piece of wire of length $\mathrm{l}$, cross-sectional area $A$ has a resistance $R$. The wire is stretched uniformly to a new length 21 . What is the resistance of the wire now?
152005
Two filaments of same length are connected first in series and then in parallel. For the same amount of main current flowing the ratio of the heat produced is
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
C When filament connected in series, $R_{\text {eq }}=R_{1}+R_{2}$ $R_{e q}=\frac{\rho L}{A}+\frac{\rho L}{A}=\frac{2 \rho L}{A}$ Power $\left(\mathrm{P}_{1}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}=\mathrm{I}^{2} \frac{2 \rho \mathrm{L}}{\mathrm{A}}$ When filament connected in parallel, $\mathrm{R}_{\text {eq }}^{\prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\frac{\mathrm{R} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}^{2}}{2 \mathrm{R}}=\frac{\mathrm{R}}{2}=\frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ Power $\left(\mathrm{P}_{2}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}^{\prime}=\mathrm{I}^{2} \frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}} =\frac{\mathrm{I}^{2} 2 \rho \mathrm{L}}{\mathrm{A}} \times \frac{2 \mathrm{~A}}{\rho \mathrm{LI}^{2}}$ $=4: 1$
151998
A $30 \mathrm{~V}, 90 \mathrm{~W}$ lamp is to be operated on a $120 \mathrm{~V}$ D.C. line. For proper glow, a resistor of ........ ohm should be connected in series with the lamp.
1 40
2 10
3 20
4 30
Explanation:
D Given, Resistance of the lamp $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{30^{2}}{90}=10 \Omega$ Current flowing the lamp $(\mathrm{I})=\frac{\mathrm{P}}{\mathrm{V}}=\frac{90}{30}=3 \mathrm{~A}$ Resistance of the line, $\mathrm{R}_{\mathrm{d}}=\frac{\mathrm{V}_{\mathrm{d}}}{\mathrm{I}}=\frac{120}{3}=40 \Omega$ If $\mathrm{R}^{\prime}$ is the resistor added, $R_{d}=R^{\prime}+R$ $R^{\prime}=R_{d}-R=40-10=30 \Omega$ Hence, for proper glow of lamp a $30 \Omega$ resistor.
Karnataka CET-2006
Current Electricity
151999
In the given circuit, the voltmeter records 5 volts. The resistance of the voltmeter in ohms is:
1 200
2 100
3 10
4 50
Explanation:
B Total potential of $10 \mathrm{~V}$ equally distributes between $50 \Omega$ and other parallel combination of $100 \Omega$ and voltmeter. Resistance of voltmeter $=\frac{100 \times \mathrm{R}}{100+\mathrm{R}}=50$ $\mathrm{R}=100 \Omega$
Karnataka CET-2005
Current Electricity
152000
Consider a rectangular slab of length $L$ and area of cross section $A$. A current $I$ is passed through it. If the length is doubled, the potential drop across the end faces
1 becomes half of the initial value
2 becomes one-fourth of the initial value
3 becomes double the initial value
4 remains same
Explanation:
C Resistance, $R=\frac{\rho l}{A}$ $\mathrm{V}=\mathrm{IR}$ $\mathrm{V}=\mathrm{I} \times \frac{\rho l}{\mathrm{~A}}$ Now, when length is doubled, $\mathrm{V}^{\prime} =\mathrm{I} \times \rho \times \frac{2 l}{\mathrm{~A}}$ $\mathrm{~V}^{\prime} =2 \mathrm{~V}$ Become double the initial value
Current Electricity
152002
A given piece of wire of length $\mathrm{l}$, cross-sectional area $A$ has a resistance $R$. The wire is stretched uniformly to a new length 21 . What is the resistance of the wire now?
152005
Two filaments of same length are connected first in series and then in parallel. For the same amount of main current flowing the ratio of the heat produced is
1 $2: 1$
2 $1: 2$
3 $4: 1$
4 $1: 4$
Explanation:
C When filament connected in series, $R_{\text {eq }}=R_{1}+R_{2}$ $R_{e q}=\frac{\rho L}{A}+\frac{\rho L}{A}=\frac{2 \rho L}{A}$ Power $\left(\mathrm{P}_{1}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}=\mathrm{I}^{2} \frac{2 \rho \mathrm{L}}{\mathrm{A}}$ When filament connected in parallel, $\mathrm{R}_{\text {eq }}^{\prime}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\frac{\mathrm{R} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}^{2}}{2 \mathrm{R}}=\frac{\mathrm{R}}{2}=\frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ Power $\left(\mathrm{P}_{2}\right)=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}^{\prime}=\mathrm{I}^{2} \frac{\rho \mathrm{L}}{2 \mathrm{~A}}$ $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}} =\frac{\mathrm{I}^{2} 2 \rho \mathrm{L}}{\mathrm{A}} \times \frac{2 \mathrm{~A}}{\rho \mathrm{LI}^{2}}$ $=4: 1$
