151904
An electric motor operates on a $50 \mathrm{~V}$ supply and a current of $12 \mathrm{~A}$. If the efficiency of the motor is $30 \%$, what is the resistance of the winding of the motor?
1 $6 \Omega$
2 $4 \Omega$
3 $2.9 \Omega$
4 $3.1 \Omega$
Explanation:
C Given, $\mathrm{V}=50 \text { Volt }$ $\mathrm{I}=12 \mathrm{~A}$ Efficiency $(\eta)=30 \%$ Power Input $\left(\mathrm{P}_{\mathrm{I}}\right)=50 \times 12$ $=600 \mathrm{~W}$ Power output $\left(\mathrm{P}_{\mathrm{o}}\right)=\frac{30}{100} \times 600=180 \mathrm{~W}$ Net power, $\mathrm{P}=\left(\mathrm{P}_{\mathrm{I}}-\mathrm{P}_{\mathrm{o}}\right)$ $=600-180=420 \mathrm{~W}$ We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $420=(12)^{2} \times \mathrm{R}$ $\mathrm{R}=\frac{420}{144}=2.9 \Omega$
Manipal UGET-2018
Current Electricity
151905
A uniform wire of resistance $9 \Omega$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $\mathrm{ABC}$. A cell of emf $2 \mathrm{~V}$ and negligible internal resistance is connected across $B$ and $C$. Potential difference across $A B$ is
1 $1 \mathrm{~V}$
2 $2 \mathrm{~V}$
3 $3 \mathrm{~V}$
4 $0.5 \mathrm{~V}$
Explanation:
A $\because \quad 3 \Omega$ and $3 \Omega$ are series In the circuit, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{6}+\frac{1}{3}=2 \Omega$ $\mathrm{R}_{\mathrm{eq}}=2 \Omega$ Now, $\mathrm{V} =\mathrm{IR}$ $2 =\mathrm{I} \times 2$ $\mathrm{I} =1 \mathrm{~A}$ $\therefore \quad \mathrm{i}_{1} =1 \times\left(\frac{3}{3+6}\right)$ $\mathrm{i}_{1} =\frac{1}{3} \mathrm{~A}$ Potential difference between $\mathrm{A}$ and $\mathrm{B}$, $V_{A}-V_{B}=\frac{1}{3} \times R$ $V_{A}-V_{B}=\frac{1}{3} \times 3$ $V_{A}-V_{B}=1 V$
Manipal UGET-2018
Current Electricity
151907
A circular coil of single turn has a resistance of $20 \Omega$. Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?
1 $5 \Omega$
2 $10 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
A Given, Resistance of circular coil of single turn $=20 \Omega$ If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel. $\therefore \quad$ Equivalent Resistance, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{10}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=5 \Omega$
NDA (I) 2017
Current Electricity
151908
On stretching a wire its length is increased by $0.2 \%$, its resistance will
1 increase by $0.1 \%$
2 decrease by $0.1 \%$
3 increase by $0.2 \%$
4 increase by $0.4 \%$
Explanation:
D $\mathrm{L}_{2}=\mathrm{L}_{1}+0.2 \%$ of $\mathrm{L}_{1}$ $\mathrm{L}_{2}=1.002 \mathrm{~L}_{1}$ $\mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}=\rho \frac{\mathrm{L}^{2}}{\mathrm{~V}}$ Where $\mathrm{A}$ is area, and $\mathrm{V}$ is volume $\mathrm{R} \propto \mathrm{L}^{2}$ Current Electricity $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right)^{2}$ $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{1.002 \mathrm{~L}_{1}}{\mathrm{~L}_{1}}\right)^{2}=(1.002)^{2}$ $\mathrm{R}_{2}=1.004 \mathrm{R}_{1}$ $R_{2}$ is increased by $0.4 \%$ of $R_{1}$
AMU-2017
Current Electricity
151910
Six wires, each of resistance $r$, are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is
1 $\mathrm{r}$
2 $2 \mathrm{r}$
3 $\frac{r}{3}$
4 $\frac{r}{2}$
Explanation:
D Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit. $\frac{1}{R}=\frac{1}{2 r}+\frac{1}{2 r}=\frac{2}{2 r}$ $R=r$ It will in parallel with outer resistance $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}=\frac{2}{\mathrm{r}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{r}}{2}$
151904
An electric motor operates on a $50 \mathrm{~V}$ supply and a current of $12 \mathrm{~A}$. If the efficiency of the motor is $30 \%$, what is the resistance of the winding of the motor?
1 $6 \Omega$
2 $4 \Omega$
3 $2.9 \Omega$
4 $3.1 \Omega$
Explanation:
C Given, $\mathrm{V}=50 \text { Volt }$ $\mathrm{I}=12 \mathrm{~A}$ Efficiency $(\eta)=30 \%$ Power Input $\left(\mathrm{P}_{\mathrm{I}}\right)=50 \times 12$ $=600 \mathrm{~W}$ Power output $\left(\mathrm{P}_{\mathrm{o}}\right)=\frac{30}{100} \times 600=180 \mathrm{~W}$ Net power, $\mathrm{P}=\left(\mathrm{P}_{\mathrm{I}}-\mathrm{P}_{\mathrm{o}}\right)$ $=600-180=420 \mathrm{~W}$ We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $420=(12)^{2} \times \mathrm{R}$ $\mathrm{R}=\frac{420}{144}=2.9 \Omega$
Manipal UGET-2018
Current Electricity
151905
A uniform wire of resistance $9 \Omega$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $\mathrm{ABC}$. A cell of emf $2 \mathrm{~V}$ and negligible internal resistance is connected across $B$ and $C$. Potential difference across $A B$ is
1 $1 \mathrm{~V}$
2 $2 \mathrm{~V}$
3 $3 \mathrm{~V}$
4 $0.5 \mathrm{~V}$
Explanation:
A $\because \quad 3 \Omega$ and $3 \Omega$ are series In the circuit, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{6}+\frac{1}{3}=2 \Omega$ $\mathrm{R}_{\mathrm{eq}}=2 \Omega$ Now, $\mathrm{V} =\mathrm{IR}$ $2 =\mathrm{I} \times 2$ $\mathrm{I} =1 \mathrm{~A}$ $\therefore \quad \mathrm{i}_{1} =1 \times\left(\frac{3}{3+6}\right)$ $\mathrm{i}_{1} =\frac{1}{3} \mathrm{~A}$ Potential difference between $\mathrm{A}$ and $\mathrm{B}$, $V_{A}-V_{B}=\frac{1}{3} \times R$ $V_{A}-V_{B}=\frac{1}{3} \times 3$ $V_{A}-V_{B}=1 V$
Manipal UGET-2018
Current Electricity
151907
A circular coil of single turn has a resistance of $20 \Omega$. Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?
1 $5 \Omega$
2 $10 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
A Given, Resistance of circular coil of single turn $=20 \Omega$ If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel. $\therefore \quad$ Equivalent Resistance, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{10}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=5 \Omega$
NDA (I) 2017
Current Electricity
151908
On stretching a wire its length is increased by $0.2 \%$, its resistance will
1 increase by $0.1 \%$
2 decrease by $0.1 \%$
3 increase by $0.2 \%$
4 increase by $0.4 \%$
Explanation:
D $\mathrm{L}_{2}=\mathrm{L}_{1}+0.2 \%$ of $\mathrm{L}_{1}$ $\mathrm{L}_{2}=1.002 \mathrm{~L}_{1}$ $\mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}=\rho \frac{\mathrm{L}^{2}}{\mathrm{~V}}$ Where $\mathrm{A}$ is area, and $\mathrm{V}$ is volume $\mathrm{R} \propto \mathrm{L}^{2}$ Current Electricity $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right)^{2}$ $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{1.002 \mathrm{~L}_{1}}{\mathrm{~L}_{1}}\right)^{2}=(1.002)^{2}$ $\mathrm{R}_{2}=1.004 \mathrm{R}_{1}$ $R_{2}$ is increased by $0.4 \%$ of $R_{1}$
AMU-2017
Current Electricity
151910
Six wires, each of resistance $r$, are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is
1 $\mathrm{r}$
2 $2 \mathrm{r}$
3 $\frac{r}{3}$
4 $\frac{r}{2}$
Explanation:
D Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit. $\frac{1}{R}=\frac{1}{2 r}+\frac{1}{2 r}=\frac{2}{2 r}$ $R=r$ It will in parallel with outer resistance $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}=\frac{2}{\mathrm{r}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{r}}{2}$
151904
An electric motor operates on a $50 \mathrm{~V}$ supply and a current of $12 \mathrm{~A}$. If the efficiency of the motor is $30 \%$, what is the resistance of the winding of the motor?
1 $6 \Omega$
2 $4 \Omega$
3 $2.9 \Omega$
4 $3.1 \Omega$
Explanation:
C Given, $\mathrm{V}=50 \text { Volt }$ $\mathrm{I}=12 \mathrm{~A}$ Efficiency $(\eta)=30 \%$ Power Input $\left(\mathrm{P}_{\mathrm{I}}\right)=50 \times 12$ $=600 \mathrm{~W}$ Power output $\left(\mathrm{P}_{\mathrm{o}}\right)=\frac{30}{100} \times 600=180 \mathrm{~W}$ Net power, $\mathrm{P}=\left(\mathrm{P}_{\mathrm{I}}-\mathrm{P}_{\mathrm{o}}\right)$ $=600-180=420 \mathrm{~W}$ We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $420=(12)^{2} \times \mathrm{R}$ $\mathrm{R}=\frac{420}{144}=2.9 \Omega$
Manipal UGET-2018
Current Electricity
151905
A uniform wire of resistance $9 \Omega$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $\mathrm{ABC}$. A cell of emf $2 \mathrm{~V}$ and negligible internal resistance is connected across $B$ and $C$. Potential difference across $A B$ is
1 $1 \mathrm{~V}$
2 $2 \mathrm{~V}$
3 $3 \mathrm{~V}$
4 $0.5 \mathrm{~V}$
Explanation:
A $\because \quad 3 \Omega$ and $3 \Omega$ are series In the circuit, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{6}+\frac{1}{3}=2 \Omega$ $\mathrm{R}_{\mathrm{eq}}=2 \Omega$ Now, $\mathrm{V} =\mathrm{IR}$ $2 =\mathrm{I} \times 2$ $\mathrm{I} =1 \mathrm{~A}$ $\therefore \quad \mathrm{i}_{1} =1 \times\left(\frac{3}{3+6}\right)$ $\mathrm{i}_{1} =\frac{1}{3} \mathrm{~A}$ Potential difference between $\mathrm{A}$ and $\mathrm{B}$, $V_{A}-V_{B}=\frac{1}{3} \times R$ $V_{A}-V_{B}=\frac{1}{3} \times 3$ $V_{A}-V_{B}=1 V$
Manipal UGET-2018
Current Electricity
151907
A circular coil of single turn has a resistance of $20 \Omega$. Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?
1 $5 \Omega$
2 $10 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
A Given, Resistance of circular coil of single turn $=20 \Omega$ If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel. $\therefore \quad$ Equivalent Resistance, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{10}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=5 \Omega$
NDA (I) 2017
Current Electricity
151908
On stretching a wire its length is increased by $0.2 \%$, its resistance will
1 increase by $0.1 \%$
2 decrease by $0.1 \%$
3 increase by $0.2 \%$
4 increase by $0.4 \%$
Explanation:
D $\mathrm{L}_{2}=\mathrm{L}_{1}+0.2 \%$ of $\mathrm{L}_{1}$ $\mathrm{L}_{2}=1.002 \mathrm{~L}_{1}$ $\mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}=\rho \frac{\mathrm{L}^{2}}{\mathrm{~V}}$ Where $\mathrm{A}$ is area, and $\mathrm{V}$ is volume $\mathrm{R} \propto \mathrm{L}^{2}$ Current Electricity $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right)^{2}$ $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{1.002 \mathrm{~L}_{1}}{\mathrm{~L}_{1}}\right)^{2}=(1.002)^{2}$ $\mathrm{R}_{2}=1.004 \mathrm{R}_{1}$ $R_{2}$ is increased by $0.4 \%$ of $R_{1}$
AMU-2017
Current Electricity
151910
Six wires, each of resistance $r$, are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is
1 $\mathrm{r}$
2 $2 \mathrm{r}$
3 $\frac{r}{3}$
4 $\frac{r}{2}$
Explanation:
D Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit. $\frac{1}{R}=\frac{1}{2 r}+\frac{1}{2 r}=\frac{2}{2 r}$ $R=r$ It will in parallel with outer resistance $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}=\frac{2}{\mathrm{r}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{r}}{2}$
151904
An electric motor operates on a $50 \mathrm{~V}$ supply and a current of $12 \mathrm{~A}$. If the efficiency of the motor is $30 \%$, what is the resistance of the winding of the motor?
1 $6 \Omega$
2 $4 \Omega$
3 $2.9 \Omega$
4 $3.1 \Omega$
Explanation:
C Given, $\mathrm{V}=50 \text { Volt }$ $\mathrm{I}=12 \mathrm{~A}$ Efficiency $(\eta)=30 \%$ Power Input $\left(\mathrm{P}_{\mathrm{I}}\right)=50 \times 12$ $=600 \mathrm{~W}$ Power output $\left(\mathrm{P}_{\mathrm{o}}\right)=\frac{30}{100} \times 600=180 \mathrm{~W}$ Net power, $\mathrm{P}=\left(\mathrm{P}_{\mathrm{I}}-\mathrm{P}_{\mathrm{o}}\right)$ $=600-180=420 \mathrm{~W}$ We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $420=(12)^{2} \times \mathrm{R}$ $\mathrm{R}=\frac{420}{144}=2.9 \Omega$
Manipal UGET-2018
Current Electricity
151905
A uniform wire of resistance $9 \Omega$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $\mathrm{ABC}$. A cell of emf $2 \mathrm{~V}$ and negligible internal resistance is connected across $B$ and $C$. Potential difference across $A B$ is
1 $1 \mathrm{~V}$
2 $2 \mathrm{~V}$
3 $3 \mathrm{~V}$
4 $0.5 \mathrm{~V}$
Explanation:
A $\because \quad 3 \Omega$ and $3 \Omega$ are series In the circuit, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{6}+\frac{1}{3}=2 \Omega$ $\mathrm{R}_{\mathrm{eq}}=2 \Omega$ Now, $\mathrm{V} =\mathrm{IR}$ $2 =\mathrm{I} \times 2$ $\mathrm{I} =1 \mathrm{~A}$ $\therefore \quad \mathrm{i}_{1} =1 \times\left(\frac{3}{3+6}\right)$ $\mathrm{i}_{1} =\frac{1}{3} \mathrm{~A}$ Potential difference between $\mathrm{A}$ and $\mathrm{B}$, $V_{A}-V_{B}=\frac{1}{3} \times R$ $V_{A}-V_{B}=\frac{1}{3} \times 3$ $V_{A}-V_{B}=1 V$
Manipal UGET-2018
Current Electricity
151907
A circular coil of single turn has a resistance of $20 \Omega$. Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?
1 $5 \Omega$
2 $10 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
A Given, Resistance of circular coil of single turn $=20 \Omega$ If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel. $\therefore \quad$ Equivalent Resistance, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{10}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=5 \Omega$
NDA (I) 2017
Current Electricity
151908
On stretching a wire its length is increased by $0.2 \%$, its resistance will
1 increase by $0.1 \%$
2 decrease by $0.1 \%$
3 increase by $0.2 \%$
4 increase by $0.4 \%$
Explanation:
D $\mathrm{L}_{2}=\mathrm{L}_{1}+0.2 \%$ of $\mathrm{L}_{1}$ $\mathrm{L}_{2}=1.002 \mathrm{~L}_{1}$ $\mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}=\rho \frac{\mathrm{L}^{2}}{\mathrm{~V}}$ Where $\mathrm{A}$ is area, and $\mathrm{V}$ is volume $\mathrm{R} \propto \mathrm{L}^{2}$ Current Electricity $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right)^{2}$ $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{1.002 \mathrm{~L}_{1}}{\mathrm{~L}_{1}}\right)^{2}=(1.002)^{2}$ $\mathrm{R}_{2}=1.004 \mathrm{R}_{1}$ $R_{2}$ is increased by $0.4 \%$ of $R_{1}$
AMU-2017
Current Electricity
151910
Six wires, each of resistance $r$, are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is
1 $\mathrm{r}$
2 $2 \mathrm{r}$
3 $\frac{r}{3}$
4 $\frac{r}{2}$
Explanation:
D Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit. $\frac{1}{R}=\frac{1}{2 r}+\frac{1}{2 r}=\frac{2}{2 r}$ $R=r$ It will in parallel with outer resistance $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}=\frac{2}{\mathrm{r}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{r}}{2}$
151904
An electric motor operates on a $50 \mathrm{~V}$ supply and a current of $12 \mathrm{~A}$. If the efficiency of the motor is $30 \%$, what is the resistance of the winding of the motor?
1 $6 \Omega$
2 $4 \Omega$
3 $2.9 \Omega$
4 $3.1 \Omega$
Explanation:
C Given, $\mathrm{V}=50 \text { Volt }$ $\mathrm{I}=12 \mathrm{~A}$ Efficiency $(\eta)=30 \%$ Power Input $\left(\mathrm{P}_{\mathrm{I}}\right)=50 \times 12$ $=600 \mathrm{~W}$ Power output $\left(\mathrm{P}_{\mathrm{o}}\right)=\frac{30}{100} \times 600=180 \mathrm{~W}$ Net power, $\mathrm{P}=\left(\mathrm{P}_{\mathrm{I}}-\mathrm{P}_{\mathrm{o}}\right)$ $=600-180=420 \mathrm{~W}$ We know that, Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $420=(12)^{2} \times \mathrm{R}$ $\mathrm{R}=\frac{420}{144}=2.9 \Omega$
Manipal UGET-2018
Current Electricity
151905
A uniform wire of resistance $9 \Omega$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $\mathrm{ABC}$. A cell of emf $2 \mathrm{~V}$ and negligible internal resistance is connected across $B$ and $C$. Potential difference across $A B$ is
1 $1 \mathrm{~V}$
2 $2 \mathrm{~V}$
3 $3 \mathrm{~V}$
4 $0.5 \mathrm{~V}$
Explanation:
A $\because \quad 3 \Omega$ and $3 \Omega$ are series In the circuit, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{6}+\frac{1}{3}=2 \Omega$ $\mathrm{R}_{\mathrm{eq}}=2 \Omega$ Now, $\mathrm{V} =\mathrm{IR}$ $2 =\mathrm{I} \times 2$ $\mathrm{I} =1 \mathrm{~A}$ $\therefore \quad \mathrm{i}_{1} =1 \times\left(\frac{3}{3+6}\right)$ $\mathrm{i}_{1} =\frac{1}{3} \mathrm{~A}$ Potential difference between $\mathrm{A}$ and $\mathrm{B}$, $V_{A}-V_{B}=\frac{1}{3} \times R$ $V_{A}-V_{B}=\frac{1}{3} \times 3$ $V_{A}-V_{B}=1 V$
Manipal UGET-2018
Current Electricity
151907
A circular coil of single turn has a resistance of $20 \Omega$. Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?
1 $5 \Omega$
2 $10 \Omega$
3 $20 \Omega$
4 $40 \Omega$
Explanation:
A Given, Resistance of circular coil of single turn $=20 \Omega$ If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel. $\therefore \quad$ Equivalent Resistance, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{10}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=5 \Omega$
NDA (I) 2017
Current Electricity
151908
On stretching a wire its length is increased by $0.2 \%$, its resistance will
1 increase by $0.1 \%$
2 decrease by $0.1 \%$
3 increase by $0.2 \%$
4 increase by $0.4 \%$
Explanation:
D $\mathrm{L}_{2}=\mathrm{L}_{1}+0.2 \%$ of $\mathrm{L}_{1}$ $\mathrm{L}_{2}=1.002 \mathrm{~L}_{1}$ $\mathrm{R}=\rho \frac{\mathrm{L}}{\mathrm{A}}=\rho \frac{\mathrm{L}^{2}}{\mathrm{~V}}$ Where $\mathrm{A}$ is area, and $\mathrm{V}$ is volume $\mathrm{R} \propto \mathrm{L}^{2}$ Current Electricity $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}\right)^{2}$ $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\left(\frac{1.002 \mathrm{~L}_{1}}{\mathrm{~L}_{1}}\right)^{2}=(1.002)^{2}$ $\mathrm{R}_{2}=1.004 \mathrm{R}_{1}$ $R_{2}$ is increased by $0.4 \%$ of $R_{1}$
AMU-2017
Current Electricity
151910
Six wires, each of resistance $r$, are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is
1 $\mathrm{r}$
2 $2 \mathrm{r}$
3 $\frac{r}{3}$
4 $\frac{r}{2}$
Explanation:
D Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit. $\frac{1}{R}=\frac{1}{2 r}+\frac{1}{2 r}=\frac{2}{2 r}$ $R=r$ It will in parallel with outer resistance $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}=\frac{2}{\mathrm{r}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{r}}{2}$
