152662
The starter motor of a car draw a current $I=$ $300 \mathrm{~A}$ from the battery of voltage $12 \mathrm{~V}$. If the car starts only after 2 minutes, what is the energy drawn from the battery?
1 $3 \mathrm{~kJ}$
2 $30 \mathrm{~kJ}$
3 $7.2 \mathrm{~kJ}$
4 $432 \mathrm{~kJ}$
Explanation:
D Given, Current (I) $=300 \mathrm{~A}$, Voltage $(\mathrm{V})=12 \mathrm{~V}$, Time $(\mathrm{t})=2 \mathrm{~min}$ So, energy drawn from the battery $(\mathrm{E})=\mathrm{VIt}$ $=12 \times 300 \times 2 \times 60$ $=432 \times 10^{3} \mathrm{~J}=432 \mathrm{~kJ}$
Manipal UGET-2019
Current Electricity
152663
Consider four circuits shown in the figure below. In which circuit power dissipated is greatest? (Neglect the internal resistance of the power supply).
1 a
2 b
3 c
4 d
Explanation:
A Let, each resistance is $\mathrm{R}$ $\mathrm{P}=\frac{\mathrm{E}^{2}}{\mathrm{R}_{\text {eq. }}}$ According to options, In option (a) - $\mathrm{R}_{\text {eq. }}=\frac{\mathrm{R} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}}{2}$ $\mathrm{P}_{1}=\frac{\mathrm{E}^{2}}{\frac{\mathrm{R}}{2}}=\frac{2 \mathrm{E}^{2}}{\mathrm{R}}$ In option (b)- $\mathrm{R}_{\text {eq. }}=\mathrm{R}+\mathrm{R}=2 \mathrm{R}$ $\mathrm{P}_{2}=\frac{\mathrm{E}^{2}}{2 \mathrm{R}}=\frac{0.5 \mathrm{E}^{2}}{\mathrm{R}}$ In option (c)- $R_{\text {eq. }}=\frac{R}{2}+R=\frac{3 R}{2}$ $P_{3}=\frac{2 E^{2}}{3 R}=0.6 \frac{E^{2}}{R}$ In option (d)- $\mathrm{R}_{\text {eq. }}=\frac{2 \mathrm{R} \times \mathrm{R}}{2 \mathrm{R}+\mathrm{R}}=\frac{2 \mathrm{R}}{3}$ $\mathrm{P}_{4}=\frac{3 \mathrm{E}^{2}}{2 \mathrm{R}}=\frac{1.5 \mathrm{E}^{2}}{\mathrm{R}}$ So, $\mathrm{P}_{1}$ is maximum.
Manipal UGET-2019
Current Electricity
152664
The resistance of a device component decreases as the current through it increases and it is described by the relation, $R=\frac{0.2 I}{I-4}$, where $I$ is the current. Determined the minimum power deliver. (Assume, $\mathrm{I}>4$ )
1 $22.4 \mathrm{~W}$
2 $18.6 \mathrm{~W}$
3 $19.8 \mathrm{~W}$
4 $21.6 \mathrm{~W}$
Explanation:
D Given, resistance $r$ of a device decreases as current 1 increases by the relation. $\mathrm{R}=\frac{0.2 \mathrm{I}}{\mathrm{I}-4}$ $\therefore$ Power of the electric device $(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $=\mathrm{I}^{2} \frac{0.2 \mathrm{I}}{\mathrm{I}-4}$ Power will be minimum if $\frac{d P}{d I}=0$ $\frac{d}{d I}\left(\frac{0.2 I^{3}}{I-4}\right)=0$ $\frac{(I-4) \times 0.2 \times 3 I^{2}-0.2 I^{3}}{(I-4)^{2}}=0$ $0.6 I^{3}-2.4 I^{2}-0.2 I^{3}=0$ $0.4 I^{3}-2.4 I^{2}=0$ $0.4 I^{2}(I-6)=0 \quad[\because I>4]$ $I-6=0$ $I=6 A$ Given, resistance $(\mathrm{R})=\frac{0.2 \mathrm{I}}{\mathrm{I}-4}=\frac{0.2 \times 6}{6-4}$ $=\frac{1.2}{2}=0.6 \Omega$ Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $=6^{2} \times 0.6=21.6 \mathrm{~W}$
TS- EAMCET-06.05.2019
Current Electricity
152666
What will be the maximum number of $60 \mathrm{~W}$ bulbs in parallel that can be turned on when a house wiring supplied with a $220 \mathrm{~V}$ supply line is protected by a 6 ampere fuse?
1 11
2 22
3 66
4 33
Explanation:
B Given, Power $(\mathrm{P})=60 \mathrm{~W}$ Fuse rating $(\mathrm{I})=6$ ampere Voltage $(\mathrm{V})=220$ Volts We know that, Power $(\mathrm{P})=\mathrm{V} \times \mathrm{I}$ $60 =220 \times \mathrm{I}$ $\therefore \quad \mathrm{I} =\frac{60}{220} \mathrm{amp}$ Current required by a single lamp. For ' $n$ ' lamps current $=\mathrm{n} \times \mathrm{I}=6 \mathrm{Amp}$ $\therefore \quad \mathrm{n}=\frac{6 \times 220}{60}$ $\mathrm{n}=22$ So, total no of bulbs required is 22 .
152662
The starter motor of a car draw a current $I=$ $300 \mathrm{~A}$ from the battery of voltage $12 \mathrm{~V}$. If the car starts only after 2 minutes, what is the energy drawn from the battery?
1 $3 \mathrm{~kJ}$
2 $30 \mathrm{~kJ}$
3 $7.2 \mathrm{~kJ}$
4 $432 \mathrm{~kJ}$
Explanation:
D Given, Current (I) $=300 \mathrm{~A}$, Voltage $(\mathrm{V})=12 \mathrm{~V}$, Time $(\mathrm{t})=2 \mathrm{~min}$ So, energy drawn from the battery $(\mathrm{E})=\mathrm{VIt}$ $=12 \times 300 \times 2 \times 60$ $=432 \times 10^{3} \mathrm{~J}=432 \mathrm{~kJ}$
Manipal UGET-2019
Current Electricity
152663
Consider four circuits shown in the figure below. In which circuit power dissipated is greatest? (Neglect the internal resistance of the power supply).
1 a
2 b
3 c
4 d
Explanation:
A Let, each resistance is $\mathrm{R}$ $\mathrm{P}=\frac{\mathrm{E}^{2}}{\mathrm{R}_{\text {eq. }}}$ According to options, In option (a) - $\mathrm{R}_{\text {eq. }}=\frac{\mathrm{R} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}}{2}$ $\mathrm{P}_{1}=\frac{\mathrm{E}^{2}}{\frac{\mathrm{R}}{2}}=\frac{2 \mathrm{E}^{2}}{\mathrm{R}}$ In option (b)- $\mathrm{R}_{\text {eq. }}=\mathrm{R}+\mathrm{R}=2 \mathrm{R}$ $\mathrm{P}_{2}=\frac{\mathrm{E}^{2}}{2 \mathrm{R}}=\frac{0.5 \mathrm{E}^{2}}{\mathrm{R}}$ In option (c)- $R_{\text {eq. }}=\frac{R}{2}+R=\frac{3 R}{2}$ $P_{3}=\frac{2 E^{2}}{3 R}=0.6 \frac{E^{2}}{R}$ In option (d)- $\mathrm{R}_{\text {eq. }}=\frac{2 \mathrm{R} \times \mathrm{R}}{2 \mathrm{R}+\mathrm{R}}=\frac{2 \mathrm{R}}{3}$ $\mathrm{P}_{4}=\frac{3 \mathrm{E}^{2}}{2 \mathrm{R}}=\frac{1.5 \mathrm{E}^{2}}{\mathrm{R}}$ So, $\mathrm{P}_{1}$ is maximum.
Manipal UGET-2019
Current Electricity
152664
The resistance of a device component decreases as the current through it increases and it is described by the relation, $R=\frac{0.2 I}{I-4}$, where $I$ is the current. Determined the minimum power deliver. (Assume, $\mathrm{I}>4$ )
1 $22.4 \mathrm{~W}$
2 $18.6 \mathrm{~W}$
3 $19.8 \mathrm{~W}$
4 $21.6 \mathrm{~W}$
Explanation:
D Given, resistance $r$ of a device decreases as current 1 increases by the relation. $\mathrm{R}=\frac{0.2 \mathrm{I}}{\mathrm{I}-4}$ $\therefore$ Power of the electric device $(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $=\mathrm{I}^{2} \frac{0.2 \mathrm{I}}{\mathrm{I}-4}$ Power will be minimum if $\frac{d P}{d I}=0$ $\frac{d}{d I}\left(\frac{0.2 I^{3}}{I-4}\right)=0$ $\frac{(I-4) \times 0.2 \times 3 I^{2}-0.2 I^{3}}{(I-4)^{2}}=0$ $0.6 I^{3}-2.4 I^{2}-0.2 I^{3}=0$ $0.4 I^{3}-2.4 I^{2}=0$ $0.4 I^{2}(I-6)=0 \quad[\because I>4]$ $I-6=0$ $I=6 A$ Given, resistance $(\mathrm{R})=\frac{0.2 \mathrm{I}}{\mathrm{I}-4}=\frac{0.2 \times 6}{6-4}$ $=\frac{1.2}{2}=0.6 \Omega$ Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $=6^{2} \times 0.6=21.6 \mathrm{~W}$
TS- EAMCET-06.05.2019
Current Electricity
152666
What will be the maximum number of $60 \mathrm{~W}$ bulbs in parallel that can be turned on when a house wiring supplied with a $220 \mathrm{~V}$ supply line is protected by a 6 ampere fuse?
1 11
2 22
3 66
4 33
Explanation:
B Given, Power $(\mathrm{P})=60 \mathrm{~W}$ Fuse rating $(\mathrm{I})=6$ ampere Voltage $(\mathrm{V})=220$ Volts We know that, Power $(\mathrm{P})=\mathrm{V} \times \mathrm{I}$ $60 =220 \times \mathrm{I}$ $\therefore \quad \mathrm{I} =\frac{60}{220} \mathrm{amp}$ Current required by a single lamp. For ' $n$ ' lamps current $=\mathrm{n} \times \mathrm{I}=6 \mathrm{Amp}$ $\therefore \quad \mathrm{n}=\frac{6 \times 220}{60}$ $\mathrm{n}=22$ So, total no of bulbs required is 22 .
152662
The starter motor of a car draw a current $I=$ $300 \mathrm{~A}$ from the battery of voltage $12 \mathrm{~V}$. If the car starts only after 2 minutes, what is the energy drawn from the battery?
1 $3 \mathrm{~kJ}$
2 $30 \mathrm{~kJ}$
3 $7.2 \mathrm{~kJ}$
4 $432 \mathrm{~kJ}$
Explanation:
D Given, Current (I) $=300 \mathrm{~A}$, Voltage $(\mathrm{V})=12 \mathrm{~V}$, Time $(\mathrm{t})=2 \mathrm{~min}$ So, energy drawn from the battery $(\mathrm{E})=\mathrm{VIt}$ $=12 \times 300 \times 2 \times 60$ $=432 \times 10^{3} \mathrm{~J}=432 \mathrm{~kJ}$
Manipal UGET-2019
Current Electricity
152663
Consider four circuits shown in the figure below. In which circuit power dissipated is greatest? (Neglect the internal resistance of the power supply).
1 a
2 b
3 c
4 d
Explanation:
A Let, each resistance is $\mathrm{R}$ $\mathrm{P}=\frac{\mathrm{E}^{2}}{\mathrm{R}_{\text {eq. }}}$ According to options, In option (a) - $\mathrm{R}_{\text {eq. }}=\frac{\mathrm{R} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}}{2}$ $\mathrm{P}_{1}=\frac{\mathrm{E}^{2}}{\frac{\mathrm{R}}{2}}=\frac{2 \mathrm{E}^{2}}{\mathrm{R}}$ In option (b)- $\mathrm{R}_{\text {eq. }}=\mathrm{R}+\mathrm{R}=2 \mathrm{R}$ $\mathrm{P}_{2}=\frac{\mathrm{E}^{2}}{2 \mathrm{R}}=\frac{0.5 \mathrm{E}^{2}}{\mathrm{R}}$ In option (c)- $R_{\text {eq. }}=\frac{R}{2}+R=\frac{3 R}{2}$ $P_{3}=\frac{2 E^{2}}{3 R}=0.6 \frac{E^{2}}{R}$ In option (d)- $\mathrm{R}_{\text {eq. }}=\frac{2 \mathrm{R} \times \mathrm{R}}{2 \mathrm{R}+\mathrm{R}}=\frac{2 \mathrm{R}}{3}$ $\mathrm{P}_{4}=\frac{3 \mathrm{E}^{2}}{2 \mathrm{R}}=\frac{1.5 \mathrm{E}^{2}}{\mathrm{R}}$ So, $\mathrm{P}_{1}$ is maximum.
Manipal UGET-2019
Current Electricity
152664
The resistance of a device component decreases as the current through it increases and it is described by the relation, $R=\frac{0.2 I}{I-4}$, where $I$ is the current. Determined the minimum power deliver. (Assume, $\mathrm{I}>4$ )
1 $22.4 \mathrm{~W}$
2 $18.6 \mathrm{~W}$
3 $19.8 \mathrm{~W}$
4 $21.6 \mathrm{~W}$
Explanation:
D Given, resistance $r$ of a device decreases as current 1 increases by the relation. $\mathrm{R}=\frac{0.2 \mathrm{I}}{\mathrm{I}-4}$ $\therefore$ Power of the electric device $(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $=\mathrm{I}^{2} \frac{0.2 \mathrm{I}}{\mathrm{I}-4}$ Power will be minimum if $\frac{d P}{d I}=0$ $\frac{d}{d I}\left(\frac{0.2 I^{3}}{I-4}\right)=0$ $\frac{(I-4) \times 0.2 \times 3 I^{2}-0.2 I^{3}}{(I-4)^{2}}=0$ $0.6 I^{3}-2.4 I^{2}-0.2 I^{3}=0$ $0.4 I^{3}-2.4 I^{2}=0$ $0.4 I^{2}(I-6)=0 \quad[\because I>4]$ $I-6=0$ $I=6 A$ Given, resistance $(\mathrm{R})=\frac{0.2 \mathrm{I}}{\mathrm{I}-4}=\frac{0.2 \times 6}{6-4}$ $=\frac{1.2}{2}=0.6 \Omega$ Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $=6^{2} \times 0.6=21.6 \mathrm{~W}$
TS- EAMCET-06.05.2019
Current Electricity
152666
What will be the maximum number of $60 \mathrm{~W}$ bulbs in parallel that can be turned on when a house wiring supplied with a $220 \mathrm{~V}$ supply line is protected by a 6 ampere fuse?
1 11
2 22
3 66
4 33
Explanation:
B Given, Power $(\mathrm{P})=60 \mathrm{~W}$ Fuse rating $(\mathrm{I})=6$ ampere Voltage $(\mathrm{V})=220$ Volts We know that, Power $(\mathrm{P})=\mathrm{V} \times \mathrm{I}$ $60 =220 \times \mathrm{I}$ $\therefore \quad \mathrm{I} =\frac{60}{220} \mathrm{amp}$ Current required by a single lamp. For ' $n$ ' lamps current $=\mathrm{n} \times \mathrm{I}=6 \mathrm{Amp}$ $\therefore \quad \mathrm{n}=\frac{6 \times 220}{60}$ $\mathrm{n}=22$ So, total no of bulbs required is 22 .
152662
The starter motor of a car draw a current $I=$ $300 \mathrm{~A}$ from the battery of voltage $12 \mathrm{~V}$. If the car starts only after 2 minutes, what is the energy drawn from the battery?
1 $3 \mathrm{~kJ}$
2 $30 \mathrm{~kJ}$
3 $7.2 \mathrm{~kJ}$
4 $432 \mathrm{~kJ}$
Explanation:
D Given, Current (I) $=300 \mathrm{~A}$, Voltage $(\mathrm{V})=12 \mathrm{~V}$, Time $(\mathrm{t})=2 \mathrm{~min}$ So, energy drawn from the battery $(\mathrm{E})=\mathrm{VIt}$ $=12 \times 300 \times 2 \times 60$ $=432 \times 10^{3} \mathrm{~J}=432 \mathrm{~kJ}$
Manipal UGET-2019
Current Electricity
152663
Consider four circuits shown in the figure below. In which circuit power dissipated is greatest? (Neglect the internal resistance of the power supply).
1 a
2 b
3 c
4 d
Explanation:
A Let, each resistance is $\mathrm{R}$ $\mathrm{P}=\frac{\mathrm{E}^{2}}{\mathrm{R}_{\text {eq. }}}$ According to options, In option (a) - $\mathrm{R}_{\text {eq. }}=\frac{\mathrm{R} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\frac{\mathrm{R}}{2}$ $\mathrm{P}_{1}=\frac{\mathrm{E}^{2}}{\frac{\mathrm{R}}{2}}=\frac{2 \mathrm{E}^{2}}{\mathrm{R}}$ In option (b)- $\mathrm{R}_{\text {eq. }}=\mathrm{R}+\mathrm{R}=2 \mathrm{R}$ $\mathrm{P}_{2}=\frac{\mathrm{E}^{2}}{2 \mathrm{R}}=\frac{0.5 \mathrm{E}^{2}}{\mathrm{R}}$ In option (c)- $R_{\text {eq. }}=\frac{R}{2}+R=\frac{3 R}{2}$ $P_{3}=\frac{2 E^{2}}{3 R}=0.6 \frac{E^{2}}{R}$ In option (d)- $\mathrm{R}_{\text {eq. }}=\frac{2 \mathrm{R} \times \mathrm{R}}{2 \mathrm{R}+\mathrm{R}}=\frac{2 \mathrm{R}}{3}$ $\mathrm{P}_{4}=\frac{3 \mathrm{E}^{2}}{2 \mathrm{R}}=\frac{1.5 \mathrm{E}^{2}}{\mathrm{R}}$ So, $\mathrm{P}_{1}$ is maximum.
Manipal UGET-2019
Current Electricity
152664
The resistance of a device component decreases as the current through it increases and it is described by the relation, $R=\frac{0.2 I}{I-4}$, where $I$ is the current. Determined the minimum power deliver. (Assume, $\mathrm{I}>4$ )
1 $22.4 \mathrm{~W}$
2 $18.6 \mathrm{~W}$
3 $19.8 \mathrm{~W}$
4 $21.6 \mathrm{~W}$
Explanation:
D Given, resistance $r$ of a device decreases as current 1 increases by the relation. $\mathrm{R}=\frac{0.2 \mathrm{I}}{\mathrm{I}-4}$ $\therefore$ Power of the electric device $(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $=\mathrm{I}^{2} \frac{0.2 \mathrm{I}}{\mathrm{I}-4}$ Power will be minimum if $\frac{d P}{d I}=0$ $\frac{d}{d I}\left(\frac{0.2 I^{3}}{I-4}\right)=0$ $\frac{(I-4) \times 0.2 \times 3 I^{2}-0.2 I^{3}}{(I-4)^{2}}=0$ $0.6 I^{3}-2.4 I^{2}-0.2 I^{3}=0$ $0.4 I^{3}-2.4 I^{2}=0$ $0.4 I^{2}(I-6)=0 \quad[\because I>4]$ $I-6=0$ $I=6 A$ Given, resistance $(\mathrm{R})=\frac{0.2 \mathrm{I}}{\mathrm{I}-4}=\frac{0.2 \times 6}{6-4}$ $=\frac{1.2}{2}=0.6 \Omega$ Power, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $=6^{2} \times 0.6=21.6 \mathrm{~W}$
TS- EAMCET-06.05.2019
Current Electricity
152666
What will be the maximum number of $60 \mathrm{~W}$ bulbs in parallel that can be turned on when a house wiring supplied with a $220 \mathrm{~V}$ supply line is protected by a 6 ampere fuse?
1 11
2 22
3 66
4 33
Explanation:
B Given, Power $(\mathrm{P})=60 \mathrm{~W}$ Fuse rating $(\mathrm{I})=6$ ampere Voltage $(\mathrm{V})=220$ Volts We know that, Power $(\mathrm{P})=\mathrm{V} \times \mathrm{I}$ $60 =220 \times \mathrm{I}$ $\therefore \quad \mathrm{I} =\frac{60}{220} \mathrm{amp}$ Current required by a single lamp. For ' $n$ ' lamps current $=\mathrm{n} \times \mathrm{I}=6 \mathrm{Amp}$ $\therefore \quad \mathrm{n}=\frac{6 \times 220}{60}$ $\mathrm{n}=22$ So, total no of bulbs required is 22 .
