152732
The power dissipated in the transmission cables carrying current $I$ and voltage $V$ is inversely proportional to
1 $\mathrm{V}$
2 $\mathrm{V}^{2}$
3 $\sqrt{\mathrm{V}}$
4 $\sqrt{\mathrm{I}}$
5 I
Explanation:
B We know that, $\mathrm{P}=\mathrm{VI}$ $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}$ Loss in transmission $=\mathrm{I}^{2} \mathrm{r}$ $=\left(\frac{\mathrm{P}}{\mathrm{V}}\right)^{2} \mathrm{r}=\left(\frac{\mathrm{P}^{2}}{\mathrm{~V}^{2}}\right) \mathrm{r} \Rightarrow \operatorname{Loss} \propto \frac{1}{\mathrm{~V}^{2}}$ So, loss in transmission is inversely proportional to $\mathrm{V}^{2}$.
Kerala CEE- 2014
Current Electricity
152733
Three identical bulbs connected in series across an accumulator consumes $20 \mathrm{~W}$ power. If the bulbs are connected in parallel to the same source, the power consumed is
1 $20 \mathrm{~W}$
2 $60 \mathrm{~W}$
3 $90 \mathrm{~W}$
4 $120 \mathrm{~W}$
5 $180 \mathrm{~W}$
Explanation:
E 'N' bulbs are identical then the total power is $\mathrm{P}_{\text {total }}=\frac{\mathrm{P}}{\mathrm{N}}$ Then in series the total power $20=\frac{P}{3}$ $P=20 \times 3=60 \mathrm{~W}$ Each bulb has power of $60 \mathrm{~W}$. In parallel combination $\mathrm{P}=\mathrm{P}_{1}+\mathrm{P}_{2}+\mathrm{P}_{3}$ $=60+60+60=180 \mathrm{~W}$
Kerala CEE 2012
Current Electricity
152734
An electric bulb rated $500 \mathrm{~W}$ at $100 \mathrm{~V}$ is used in a circuit having a $200 \mathrm{~V}$ supply. The resistance $R$ that must be put in series with the bulb, so that the bulb draws $500 \mathrm{~W}$ is
1 $10 \Omega$
2 $15 \Omega$
3 $2.5 \Omega$
4 $25 \Omega$
5 $20 \Omega$
Explanation:
E We know that, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance of bulb, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $=\frac{100 \times 100}{500}=20 \Omega$ From maximum power transfer theorem, Maximum power will be transfer when, $\mathrm{R}=\mathrm{R}_{\text {internal }}$ $\mathrm{R}=20 \Omega$
Kerala CEE - 2010
Current Electricity
152735
A heater of $220 \mathrm{~V}$ heats a volume of water in 5 min. The same heater when connected to $110 \mathrm{~V}$ heats the same volume of water in (minute)
1 5
2 20
3 10
4 2.5
5 1.25
Explanation:
B We know that, Heat produced the heater $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \times \mathrm{t}$ $220 \mathrm{~V}$ heat produced, $\mathrm{H}_{1}=\frac{(220)^{2}}{\mathrm{R}} \times 5$ $110 \mathrm{~V}$ heat produced $\mathrm{H}_{2}=\frac{(110)^{2}}{\mathrm{R}} \times \mathrm{t}$ $\mathrm{H}_{1}=\mathrm{H}_{2}$ $\frac{110 \times 110}{\mathrm{R}} \mathrm{t}=\frac{220 \times 220 \times 5}{\mathrm{R}}$ $\mathrm{t}=2 \times 2 \times 5$ $\mathrm{t}=20 \text { min }$
152732
The power dissipated in the transmission cables carrying current $I$ and voltage $V$ is inversely proportional to
1 $\mathrm{V}$
2 $\mathrm{V}^{2}$
3 $\sqrt{\mathrm{V}}$
4 $\sqrt{\mathrm{I}}$
5 I
Explanation:
B We know that, $\mathrm{P}=\mathrm{VI}$ $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}$ Loss in transmission $=\mathrm{I}^{2} \mathrm{r}$ $=\left(\frac{\mathrm{P}}{\mathrm{V}}\right)^{2} \mathrm{r}=\left(\frac{\mathrm{P}^{2}}{\mathrm{~V}^{2}}\right) \mathrm{r} \Rightarrow \operatorname{Loss} \propto \frac{1}{\mathrm{~V}^{2}}$ So, loss in transmission is inversely proportional to $\mathrm{V}^{2}$.
Kerala CEE- 2014
Current Electricity
152733
Three identical bulbs connected in series across an accumulator consumes $20 \mathrm{~W}$ power. If the bulbs are connected in parallel to the same source, the power consumed is
1 $20 \mathrm{~W}$
2 $60 \mathrm{~W}$
3 $90 \mathrm{~W}$
4 $120 \mathrm{~W}$
5 $180 \mathrm{~W}$
Explanation:
E 'N' bulbs are identical then the total power is $\mathrm{P}_{\text {total }}=\frac{\mathrm{P}}{\mathrm{N}}$ Then in series the total power $20=\frac{P}{3}$ $P=20 \times 3=60 \mathrm{~W}$ Each bulb has power of $60 \mathrm{~W}$. In parallel combination $\mathrm{P}=\mathrm{P}_{1}+\mathrm{P}_{2}+\mathrm{P}_{3}$ $=60+60+60=180 \mathrm{~W}$
Kerala CEE 2012
Current Electricity
152734
An electric bulb rated $500 \mathrm{~W}$ at $100 \mathrm{~V}$ is used in a circuit having a $200 \mathrm{~V}$ supply. The resistance $R$ that must be put in series with the bulb, so that the bulb draws $500 \mathrm{~W}$ is
1 $10 \Omega$
2 $15 \Omega$
3 $2.5 \Omega$
4 $25 \Omega$
5 $20 \Omega$
Explanation:
E We know that, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance of bulb, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $=\frac{100 \times 100}{500}=20 \Omega$ From maximum power transfer theorem, Maximum power will be transfer when, $\mathrm{R}=\mathrm{R}_{\text {internal }}$ $\mathrm{R}=20 \Omega$
Kerala CEE - 2010
Current Electricity
152735
A heater of $220 \mathrm{~V}$ heats a volume of water in 5 min. The same heater when connected to $110 \mathrm{~V}$ heats the same volume of water in (minute)
1 5
2 20
3 10
4 2.5
5 1.25
Explanation:
B We know that, Heat produced the heater $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \times \mathrm{t}$ $220 \mathrm{~V}$ heat produced, $\mathrm{H}_{1}=\frac{(220)^{2}}{\mathrm{R}} \times 5$ $110 \mathrm{~V}$ heat produced $\mathrm{H}_{2}=\frac{(110)^{2}}{\mathrm{R}} \times \mathrm{t}$ $\mathrm{H}_{1}=\mathrm{H}_{2}$ $\frac{110 \times 110}{\mathrm{R}} \mathrm{t}=\frac{220 \times 220 \times 5}{\mathrm{R}}$ $\mathrm{t}=2 \times 2 \times 5$ $\mathrm{t}=20 \text { min }$
152732
The power dissipated in the transmission cables carrying current $I$ and voltage $V$ is inversely proportional to
1 $\mathrm{V}$
2 $\mathrm{V}^{2}$
3 $\sqrt{\mathrm{V}}$
4 $\sqrt{\mathrm{I}}$
5 I
Explanation:
B We know that, $\mathrm{P}=\mathrm{VI}$ $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}$ Loss in transmission $=\mathrm{I}^{2} \mathrm{r}$ $=\left(\frac{\mathrm{P}}{\mathrm{V}}\right)^{2} \mathrm{r}=\left(\frac{\mathrm{P}^{2}}{\mathrm{~V}^{2}}\right) \mathrm{r} \Rightarrow \operatorname{Loss} \propto \frac{1}{\mathrm{~V}^{2}}$ So, loss in transmission is inversely proportional to $\mathrm{V}^{2}$.
Kerala CEE- 2014
Current Electricity
152733
Three identical bulbs connected in series across an accumulator consumes $20 \mathrm{~W}$ power. If the bulbs are connected in parallel to the same source, the power consumed is
1 $20 \mathrm{~W}$
2 $60 \mathrm{~W}$
3 $90 \mathrm{~W}$
4 $120 \mathrm{~W}$
5 $180 \mathrm{~W}$
Explanation:
E 'N' bulbs are identical then the total power is $\mathrm{P}_{\text {total }}=\frac{\mathrm{P}}{\mathrm{N}}$ Then in series the total power $20=\frac{P}{3}$ $P=20 \times 3=60 \mathrm{~W}$ Each bulb has power of $60 \mathrm{~W}$. In parallel combination $\mathrm{P}=\mathrm{P}_{1}+\mathrm{P}_{2}+\mathrm{P}_{3}$ $=60+60+60=180 \mathrm{~W}$
Kerala CEE 2012
Current Electricity
152734
An electric bulb rated $500 \mathrm{~W}$ at $100 \mathrm{~V}$ is used in a circuit having a $200 \mathrm{~V}$ supply. The resistance $R$ that must be put in series with the bulb, so that the bulb draws $500 \mathrm{~W}$ is
1 $10 \Omega$
2 $15 \Omega$
3 $2.5 \Omega$
4 $25 \Omega$
5 $20 \Omega$
Explanation:
E We know that, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance of bulb, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $=\frac{100 \times 100}{500}=20 \Omega$ From maximum power transfer theorem, Maximum power will be transfer when, $\mathrm{R}=\mathrm{R}_{\text {internal }}$ $\mathrm{R}=20 \Omega$
Kerala CEE - 2010
Current Electricity
152735
A heater of $220 \mathrm{~V}$ heats a volume of water in 5 min. The same heater when connected to $110 \mathrm{~V}$ heats the same volume of water in (minute)
1 5
2 20
3 10
4 2.5
5 1.25
Explanation:
B We know that, Heat produced the heater $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \times \mathrm{t}$ $220 \mathrm{~V}$ heat produced, $\mathrm{H}_{1}=\frac{(220)^{2}}{\mathrm{R}} \times 5$ $110 \mathrm{~V}$ heat produced $\mathrm{H}_{2}=\frac{(110)^{2}}{\mathrm{R}} \times \mathrm{t}$ $\mathrm{H}_{1}=\mathrm{H}_{2}$ $\frac{110 \times 110}{\mathrm{R}} \mathrm{t}=\frac{220 \times 220 \times 5}{\mathrm{R}}$ $\mathrm{t}=2 \times 2 \times 5$ $\mathrm{t}=20 \text { min }$
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Current Electricity
152732
The power dissipated in the transmission cables carrying current $I$ and voltage $V$ is inversely proportional to
1 $\mathrm{V}$
2 $\mathrm{V}^{2}$
3 $\sqrt{\mathrm{V}}$
4 $\sqrt{\mathrm{I}}$
5 I
Explanation:
B We know that, $\mathrm{P}=\mathrm{VI}$ $\mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}$ Loss in transmission $=\mathrm{I}^{2} \mathrm{r}$ $=\left(\frac{\mathrm{P}}{\mathrm{V}}\right)^{2} \mathrm{r}=\left(\frac{\mathrm{P}^{2}}{\mathrm{~V}^{2}}\right) \mathrm{r} \Rightarrow \operatorname{Loss} \propto \frac{1}{\mathrm{~V}^{2}}$ So, loss in transmission is inversely proportional to $\mathrm{V}^{2}$.
Kerala CEE- 2014
Current Electricity
152733
Three identical bulbs connected in series across an accumulator consumes $20 \mathrm{~W}$ power. If the bulbs are connected in parallel to the same source, the power consumed is
1 $20 \mathrm{~W}$
2 $60 \mathrm{~W}$
3 $90 \mathrm{~W}$
4 $120 \mathrm{~W}$
5 $180 \mathrm{~W}$
Explanation:
E 'N' bulbs are identical then the total power is $\mathrm{P}_{\text {total }}=\frac{\mathrm{P}}{\mathrm{N}}$ Then in series the total power $20=\frac{P}{3}$ $P=20 \times 3=60 \mathrm{~W}$ Each bulb has power of $60 \mathrm{~W}$. In parallel combination $\mathrm{P}=\mathrm{P}_{1}+\mathrm{P}_{2}+\mathrm{P}_{3}$ $=60+60+60=180 \mathrm{~W}$
Kerala CEE 2012
Current Electricity
152734
An electric bulb rated $500 \mathrm{~W}$ at $100 \mathrm{~V}$ is used in a circuit having a $200 \mathrm{~V}$ supply. The resistance $R$ that must be put in series with the bulb, so that the bulb draws $500 \mathrm{~W}$ is
1 $10 \Omega$
2 $15 \Omega$
3 $2.5 \Omega$
4 $25 \Omega$
5 $20 \Omega$
Explanation:
E We know that, $\text { Power, } \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance of bulb, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $=\frac{100 \times 100}{500}=20 \Omega$ From maximum power transfer theorem, Maximum power will be transfer when, $\mathrm{R}=\mathrm{R}_{\text {internal }}$ $\mathrm{R}=20 \Omega$
Kerala CEE - 2010
Current Electricity
152735
A heater of $220 \mathrm{~V}$ heats a volume of water in 5 min. The same heater when connected to $110 \mathrm{~V}$ heats the same volume of water in (minute)
1 5
2 20
3 10
4 2.5
5 1.25
Explanation:
B We know that, Heat produced the heater $\mathrm{H}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \times \mathrm{t}$ $220 \mathrm{~V}$ heat produced, $\mathrm{H}_{1}=\frac{(220)^{2}}{\mathrm{R}} \times 5$ $110 \mathrm{~V}$ heat produced $\mathrm{H}_{2}=\frac{(110)^{2}}{\mathrm{R}} \times \mathrm{t}$ $\mathrm{H}_{1}=\mathrm{H}_{2}$ $\frac{110 \times 110}{\mathrm{R}} \mathrm{t}=\frac{220 \times 220 \times 5}{\mathrm{R}}$ $\mathrm{t}=2 \times 2 \times 5$ $\mathrm{t}=20 \text { min }$
