152771
A wire when connected to $220 \mathrm{~V}$ mains supply has power dissipation $P_{1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$. Then $P_{2}: P_{1}$ is
1 1
2 4
3 2
4 3
Explanation:
B We know that, $\text { Power, } \mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ When connected in parallel $\mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{R} / 2) \times(\mathrm{R} / 2)}{\frac{\mathrm{R}}{2}+\frac{\mathrm{R}}{2}}=\frac{\mathrm{R}}{4}$ $\mathrm{P}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{R} / 4}=4 \frac{\mathrm{V}^{2}}{\mathrm{R}}=4 \mathrm{P}_{1}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=4$
AMU-2015
Current Electricity
152772
On a heater coil it is written that, $250 \mathrm{~V}, 500$ $W$. What is the resistance of this coil?
1 $62.5 \Omega$
2 $100 \Omega$
3 $200 \Omega$
4 $125 \Omega$
Explanation:
D Suppose that, $\mathrm{V}=250$ volt $\mathrm{P}=500 \mathrm{~W}$ Power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{250 \times 250}{500}$ $\mathrm{R}=125 \Omega$
CG PET- 2015
Current Electricity
152773
A constant potential difference is applied across the ends of a wire. Which one of the following operations will reduce the rate of heat generation to half?
152774
Two bulbs when connected in parallel to a source take $60 \mathrm{~W}$ each, the total power consumed, when they are connected in series with the same source is
1 $15 \mathrm{~W}$
2 $30 \mathrm{~W}$
3 $60 \mathrm{~W}$
4 $120 \mathrm{~W}$
Explanation:
B Given, Two bulb of Power, $\mathrm{P}_{1}=\mathrm{P}_{2}=60 \mathrm{~W}$ We know that, $\mathrm{P}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$ So, $\quad \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{1}}$ and $\quad \mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}$ Now, when both bulb connected in series, $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ $R_{\text {eq }}=\frac{V^{2}}{P_{1}}+\frac{V^{2}}{P_{2}}$ $R_{\text {eq }}=\frac{V^{2}}{60}+\frac{V^{2}}{60}=\frac{2 V^{2}}{60}=\frac{V^{2}}{30}$ Put the value of $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{30}$ in equation (i), $\mathrm{P}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{V}^{2}}{30}}$ $\mathrm{P}_{\mathrm{eq}}=30 \mathrm{~W}$
152771
A wire when connected to $220 \mathrm{~V}$ mains supply has power dissipation $P_{1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$. Then $P_{2}: P_{1}$ is
1 1
2 4
3 2
4 3
Explanation:
B We know that, $\text { Power, } \mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ When connected in parallel $\mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{R} / 2) \times(\mathrm{R} / 2)}{\frac{\mathrm{R}}{2}+\frac{\mathrm{R}}{2}}=\frac{\mathrm{R}}{4}$ $\mathrm{P}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{R} / 4}=4 \frac{\mathrm{V}^{2}}{\mathrm{R}}=4 \mathrm{P}_{1}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=4$
AMU-2015
Current Electricity
152772
On a heater coil it is written that, $250 \mathrm{~V}, 500$ $W$. What is the resistance of this coil?
1 $62.5 \Omega$
2 $100 \Omega$
3 $200 \Omega$
4 $125 \Omega$
Explanation:
D Suppose that, $\mathrm{V}=250$ volt $\mathrm{P}=500 \mathrm{~W}$ Power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{250 \times 250}{500}$ $\mathrm{R}=125 \Omega$
CG PET- 2015
Current Electricity
152773
A constant potential difference is applied across the ends of a wire. Which one of the following operations will reduce the rate of heat generation to half?
152774
Two bulbs when connected in parallel to a source take $60 \mathrm{~W}$ each, the total power consumed, when they are connected in series with the same source is
1 $15 \mathrm{~W}$
2 $30 \mathrm{~W}$
3 $60 \mathrm{~W}$
4 $120 \mathrm{~W}$
Explanation:
B Given, Two bulb of Power, $\mathrm{P}_{1}=\mathrm{P}_{2}=60 \mathrm{~W}$ We know that, $\mathrm{P}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$ So, $\quad \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{1}}$ and $\quad \mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}$ Now, when both bulb connected in series, $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ $R_{\text {eq }}=\frac{V^{2}}{P_{1}}+\frac{V^{2}}{P_{2}}$ $R_{\text {eq }}=\frac{V^{2}}{60}+\frac{V^{2}}{60}=\frac{2 V^{2}}{60}=\frac{V^{2}}{30}$ Put the value of $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{30}$ in equation (i), $\mathrm{P}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{V}^{2}}{30}}$ $\mathrm{P}_{\mathrm{eq}}=30 \mathrm{~W}$
152771
A wire when connected to $220 \mathrm{~V}$ mains supply has power dissipation $P_{1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$. Then $P_{2}: P_{1}$ is
1 1
2 4
3 2
4 3
Explanation:
B We know that, $\text { Power, } \mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ When connected in parallel $\mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{R} / 2) \times(\mathrm{R} / 2)}{\frac{\mathrm{R}}{2}+\frac{\mathrm{R}}{2}}=\frac{\mathrm{R}}{4}$ $\mathrm{P}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{R} / 4}=4 \frac{\mathrm{V}^{2}}{\mathrm{R}}=4 \mathrm{P}_{1}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=4$
AMU-2015
Current Electricity
152772
On a heater coil it is written that, $250 \mathrm{~V}, 500$ $W$. What is the resistance of this coil?
1 $62.5 \Omega$
2 $100 \Omega$
3 $200 \Omega$
4 $125 \Omega$
Explanation:
D Suppose that, $\mathrm{V}=250$ volt $\mathrm{P}=500 \mathrm{~W}$ Power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{250 \times 250}{500}$ $\mathrm{R}=125 \Omega$
CG PET- 2015
Current Electricity
152773
A constant potential difference is applied across the ends of a wire. Which one of the following operations will reduce the rate of heat generation to half?
152774
Two bulbs when connected in parallel to a source take $60 \mathrm{~W}$ each, the total power consumed, when they are connected in series with the same source is
1 $15 \mathrm{~W}$
2 $30 \mathrm{~W}$
3 $60 \mathrm{~W}$
4 $120 \mathrm{~W}$
Explanation:
B Given, Two bulb of Power, $\mathrm{P}_{1}=\mathrm{P}_{2}=60 \mathrm{~W}$ We know that, $\mathrm{P}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$ So, $\quad \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{1}}$ and $\quad \mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}$ Now, when both bulb connected in series, $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ $R_{\text {eq }}=\frac{V^{2}}{P_{1}}+\frac{V^{2}}{P_{2}}$ $R_{\text {eq }}=\frac{V^{2}}{60}+\frac{V^{2}}{60}=\frac{2 V^{2}}{60}=\frac{V^{2}}{30}$ Put the value of $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{30}$ in equation (i), $\mathrm{P}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{V}^{2}}{30}}$ $\mathrm{P}_{\mathrm{eq}}=30 \mathrm{~W}$
152771
A wire when connected to $220 \mathrm{~V}$ mains supply has power dissipation $P_{1}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is $P_{2}$. Then $P_{2}: P_{1}$ is
1 1
2 4
3 2
4 3
Explanation:
B We know that, $\text { Power, } \mathrm{P}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ When connected in parallel $\mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{R} / 2) \times(\mathrm{R} / 2)}{\frac{\mathrm{R}}{2}+\frac{\mathrm{R}}{2}}=\frac{\mathrm{R}}{4}$ $\mathrm{P}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{R} / 4}=4 \frac{\mathrm{V}^{2}}{\mathrm{R}}=4 \mathrm{P}_{1}$ $\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=4$
AMU-2015
Current Electricity
152772
On a heater coil it is written that, $250 \mathrm{~V}, 500$ $W$. What is the resistance of this coil?
1 $62.5 \Omega$
2 $100 \Omega$
3 $200 \Omega$
4 $125 \Omega$
Explanation:
D Suppose that, $\mathrm{V}=250$ volt $\mathrm{P}=500 \mathrm{~W}$ Power, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ Resistance, $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $\mathrm{R}=\frac{250 \times 250}{500}$ $\mathrm{R}=125 \Omega$
CG PET- 2015
Current Electricity
152773
A constant potential difference is applied across the ends of a wire. Which one of the following operations will reduce the rate of heat generation to half?
152774
Two bulbs when connected in parallel to a source take $60 \mathrm{~W}$ each, the total power consumed, when they are connected in series with the same source is
1 $15 \mathrm{~W}$
2 $30 \mathrm{~W}$
3 $60 \mathrm{~W}$
4 $120 \mathrm{~W}$
Explanation:
B Given, Two bulb of Power, $\mathrm{P}_{1}=\mathrm{P}_{2}=60 \mathrm{~W}$ We know that, $\mathrm{P}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$ So, $\quad \mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{1}}$ and $\quad \mathrm{R}_{2}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}$ Now, when both bulb connected in series, $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}$ $R_{\text {eq }}=\frac{V^{2}}{P_{1}}+\frac{V^{2}}{P_{2}}$ $R_{\text {eq }}=\frac{V^{2}}{60}+\frac{V^{2}}{60}=\frac{2 V^{2}}{60}=\frac{V^{2}}{30}$ Put the value of $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{30}$ in equation (i), $\mathrm{P}_{\mathrm{eq}}=\frac{\mathrm{V}^{2}}{\frac{\mathrm{V}^{2}}{30}}$ $\mathrm{P}_{\mathrm{eq}}=30 \mathrm{~W}$
