NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152649
The power dissipated by a resistance $R$, carrying a sinusoidal $\mathrm{AC}$ current of peak value $I_{P}$ is
1 $I_{P}^{2} R \cos \theta$
2 $\frac{1}{2} I_{P}^{2} R$
3 $\frac{4}{\pi} I_{P}^{2} R$
4 $\frac{1}{\pi} I_{P}^{2} R$
Explanation:
B $\text { Resistance }=\mathrm{R}$ Peak current $\left(\mathrm{I}_{\mathrm{P}}\right)=\sqrt{2} \mathrm{I}_{\mathrm{rms}}$ Peak voltage $\left(\mathrm{V}_{\mathrm{P}}\right)=\sqrt{2} \mathrm{~V}_{\mathrm{rms}}$ Power $=V_{\text {rms }} \times I_{\text {rms }}$ $=\frac{\mathrm{V}_{\mathrm{P}}}{\sqrt{2}} \times \frac{\mathrm{I}_{\mathrm{P}}}{\sqrt{2}}$ $=\frac{1}{2}\left(\mathrm{I}_{\mathrm{P}}\right)\left(\mathrm{I}_{\mathrm{P}}\right) \mathrm{R} \quad\left[\because \mathrm{V}_{\mathrm{P}}=\mathrm{I}_{\mathrm{P}} \mathrm{R}\right]$ $\mathrm{P}=\frac{1}{2} \mathrm{I}_{\mathrm{P}}^{2} \mathrm{R}$
AP EAMCET-25.09.2020
Current Electricity
152650
Two resistances $A(24 \Omega)$ and $B(6 \Omega)$ are connected in series to a d.c. supply. The ratio of heat generated $W_{A}: W_{B}$ Will be
1 $1: 16$
2 $16: 1$
3 $4: 1$
4 $1: 4$
Explanation:
C We know that, Heat generation $=I^{2} R t$ For resistance $\mathrm{A}$, Heat generation in $\mathrm{A}\left(\mathrm{W}_{\mathrm{A}}\right)=\mathrm{I}_{1}{ }^{2} \times 24 \times \mathrm{t}$ For resistance B, Heat generation in $\mathrm{B}\left(\mathrm{W}_{\mathrm{B}}\right)=\mathrm{I}_{2}{ }^{2} \times 6 \times \mathrm{t}$ $\mathrm{I}_{1}=\mathrm{I}_{2}$ because both are in series. The ratios of $\mathrm{W}_{\mathrm{A}}$ and $\mathrm{W}_{\mathrm{B}}$ is- $\frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{\mathrm{I}_{1}^{2} \times 24 \times \mathrm{t}}{\mathrm{I}_{2}^{2} \times 6 \times \mathrm{t}}$ $\frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{4}{1}$ $\mathrm{~W}_{\mathrm{A}}: \mathrm{W}_{\mathrm{B}}=4: 1$
Tripura-2020
Current Electricity
152651
A bulb of $100 \mathrm{~W}$ rating is connected with $220 \mathrm{~V}$ supply. The resistance of bulb is
1 $484 \Omega$
2 $484{\Omega \mathrm{m}^{-1}}$
3 $2.2 \Omega$
4 $2.2 \times 10^{-3} \Omega \mathrm{m}^{-1}$
Explanation:
A Given, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{~V}=220 \mathrm{~V}$ The resistance of the bulb, $\mathrm{R} =\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $=\frac{220 \times 220}{100}=484 \Omega$
GUJCET 18.04.2022
Current Electricity
152652
An electric current of 2 A passes through a wire of resistance $25 \Omega$. How much heat will be generated in $1 \mathrm{~min}$ ?
1 $6 \times 10^{3} \mathrm{~J}$
2 $3.6 \times 10^{3} \mathrm{~J}$
3 $0.6 \times 10^{3} \mathrm{~J}$
4 $0.36 \times 10^{3} \mathrm{~J}$
Explanation:
A Given, Current through the wire (I) $=2 \mathrm{~A}$ Resistance of the wire $(\mathrm{R})=25 \Omega$ Time $=1 \mathrm{~min}=60 \mathrm{sec}$ We know that, Joule's heating effect, heat developed across the wire, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\mathrm{H}=(2)^{2} \times 25 \times 60$ $\mathrm{H}=4 \times 25 \times 60$ $\mathrm{H}=6000 \mathrm{~J}$ $\mathrm{H}=6 \times 10^{3} \mathrm{~J}$
152649
The power dissipated by a resistance $R$, carrying a sinusoidal $\mathrm{AC}$ current of peak value $I_{P}$ is
1 $I_{P}^{2} R \cos \theta$
2 $\frac{1}{2} I_{P}^{2} R$
3 $\frac{4}{\pi} I_{P}^{2} R$
4 $\frac{1}{\pi} I_{P}^{2} R$
Explanation:
B $\text { Resistance }=\mathrm{R}$ Peak current $\left(\mathrm{I}_{\mathrm{P}}\right)=\sqrt{2} \mathrm{I}_{\mathrm{rms}}$ Peak voltage $\left(\mathrm{V}_{\mathrm{P}}\right)=\sqrt{2} \mathrm{~V}_{\mathrm{rms}}$ Power $=V_{\text {rms }} \times I_{\text {rms }}$ $=\frac{\mathrm{V}_{\mathrm{P}}}{\sqrt{2}} \times \frac{\mathrm{I}_{\mathrm{P}}}{\sqrt{2}}$ $=\frac{1}{2}\left(\mathrm{I}_{\mathrm{P}}\right)\left(\mathrm{I}_{\mathrm{P}}\right) \mathrm{R} \quad\left[\because \mathrm{V}_{\mathrm{P}}=\mathrm{I}_{\mathrm{P}} \mathrm{R}\right]$ $\mathrm{P}=\frac{1}{2} \mathrm{I}_{\mathrm{P}}^{2} \mathrm{R}$
AP EAMCET-25.09.2020
Current Electricity
152650
Two resistances $A(24 \Omega)$ and $B(6 \Omega)$ are connected in series to a d.c. supply. The ratio of heat generated $W_{A}: W_{B}$ Will be
1 $1: 16$
2 $16: 1$
3 $4: 1$
4 $1: 4$
Explanation:
C We know that, Heat generation $=I^{2} R t$ For resistance $\mathrm{A}$, Heat generation in $\mathrm{A}\left(\mathrm{W}_{\mathrm{A}}\right)=\mathrm{I}_{1}{ }^{2} \times 24 \times \mathrm{t}$ For resistance B, Heat generation in $\mathrm{B}\left(\mathrm{W}_{\mathrm{B}}\right)=\mathrm{I}_{2}{ }^{2} \times 6 \times \mathrm{t}$ $\mathrm{I}_{1}=\mathrm{I}_{2}$ because both are in series. The ratios of $\mathrm{W}_{\mathrm{A}}$ and $\mathrm{W}_{\mathrm{B}}$ is- $\frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{\mathrm{I}_{1}^{2} \times 24 \times \mathrm{t}}{\mathrm{I}_{2}^{2} \times 6 \times \mathrm{t}}$ $\frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{4}{1}$ $\mathrm{~W}_{\mathrm{A}}: \mathrm{W}_{\mathrm{B}}=4: 1$
Tripura-2020
Current Electricity
152651
A bulb of $100 \mathrm{~W}$ rating is connected with $220 \mathrm{~V}$ supply. The resistance of bulb is
1 $484 \Omega$
2 $484{\Omega \mathrm{m}^{-1}}$
3 $2.2 \Omega$
4 $2.2 \times 10^{-3} \Omega \mathrm{m}^{-1}$
Explanation:
A Given, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{~V}=220 \mathrm{~V}$ The resistance of the bulb, $\mathrm{R} =\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $=\frac{220 \times 220}{100}=484 \Omega$
GUJCET 18.04.2022
Current Electricity
152652
An electric current of 2 A passes through a wire of resistance $25 \Omega$. How much heat will be generated in $1 \mathrm{~min}$ ?
1 $6 \times 10^{3} \mathrm{~J}$
2 $3.6 \times 10^{3} \mathrm{~J}$
3 $0.6 \times 10^{3} \mathrm{~J}$
4 $0.36 \times 10^{3} \mathrm{~J}$
Explanation:
A Given, Current through the wire (I) $=2 \mathrm{~A}$ Resistance of the wire $(\mathrm{R})=25 \Omega$ Time $=1 \mathrm{~min}=60 \mathrm{sec}$ We know that, Joule's heating effect, heat developed across the wire, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\mathrm{H}=(2)^{2} \times 25 \times 60$ $\mathrm{H}=4 \times 25 \times 60$ $\mathrm{H}=6000 \mathrm{~J}$ $\mathrm{H}=6 \times 10^{3} \mathrm{~J}$
152649
The power dissipated by a resistance $R$, carrying a sinusoidal $\mathrm{AC}$ current of peak value $I_{P}$ is
1 $I_{P}^{2} R \cos \theta$
2 $\frac{1}{2} I_{P}^{2} R$
3 $\frac{4}{\pi} I_{P}^{2} R$
4 $\frac{1}{\pi} I_{P}^{2} R$
Explanation:
B $\text { Resistance }=\mathrm{R}$ Peak current $\left(\mathrm{I}_{\mathrm{P}}\right)=\sqrt{2} \mathrm{I}_{\mathrm{rms}}$ Peak voltage $\left(\mathrm{V}_{\mathrm{P}}\right)=\sqrt{2} \mathrm{~V}_{\mathrm{rms}}$ Power $=V_{\text {rms }} \times I_{\text {rms }}$ $=\frac{\mathrm{V}_{\mathrm{P}}}{\sqrt{2}} \times \frac{\mathrm{I}_{\mathrm{P}}}{\sqrt{2}}$ $=\frac{1}{2}\left(\mathrm{I}_{\mathrm{P}}\right)\left(\mathrm{I}_{\mathrm{P}}\right) \mathrm{R} \quad\left[\because \mathrm{V}_{\mathrm{P}}=\mathrm{I}_{\mathrm{P}} \mathrm{R}\right]$ $\mathrm{P}=\frac{1}{2} \mathrm{I}_{\mathrm{P}}^{2} \mathrm{R}$
AP EAMCET-25.09.2020
Current Electricity
152650
Two resistances $A(24 \Omega)$ and $B(6 \Omega)$ are connected in series to a d.c. supply. The ratio of heat generated $W_{A}: W_{B}$ Will be
1 $1: 16$
2 $16: 1$
3 $4: 1$
4 $1: 4$
Explanation:
C We know that, Heat generation $=I^{2} R t$ For resistance $\mathrm{A}$, Heat generation in $\mathrm{A}\left(\mathrm{W}_{\mathrm{A}}\right)=\mathrm{I}_{1}{ }^{2} \times 24 \times \mathrm{t}$ For resistance B, Heat generation in $\mathrm{B}\left(\mathrm{W}_{\mathrm{B}}\right)=\mathrm{I}_{2}{ }^{2} \times 6 \times \mathrm{t}$ $\mathrm{I}_{1}=\mathrm{I}_{2}$ because both are in series. The ratios of $\mathrm{W}_{\mathrm{A}}$ and $\mathrm{W}_{\mathrm{B}}$ is- $\frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{\mathrm{I}_{1}^{2} \times 24 \times \mathrm{t}}{\mathrm{I}_{2}^{2} \times 6 \times \mathrm{t}}$ $\frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{4}{1}$ $\mathrm{~W}_{\mathrm{A}}: \mathrm{W}_{\mathrm{B}}=4: 1$
Tripura-2020
Current Electricity
152651
A bulb of $100 \mathrm{~W}$ rating is connected with $220 \mathrm{~V}$ supply. The resistance of bulb is
1 $484 \Omega$
2 $484{\Omega \mathrm{m}^{-1}}$
3 $2.2 \Omega$
4 $2.2 \times 10^{-3} \Omega \mathrm{m}^{-1}$
Explanation:
A Given, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{~V}=220 \mathrm{~V}$ The resistance of the bulb, $\mathrm{R} =\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $=\frac{220 \times 220}{100}=484 \Omega$
GUJCET 18.04.2022
Current Electricity
152652
An electric current of 2 A passes through a wire of resistance $25 \Omega$. How much heat will be generated in $1 \mathrm{~min}$ ?
1 $6 \times 10^{3} \mathrm{~J}$
2 $3.6 \times 10^{3} \mathrm{~J}$
3 $0.6 \times 10^{3} \mathrm{~J}$
4 $0.36 \times 10^{3} \mathrm{~J}$
Explanation:
A Given, Current through the wire (I) $=2 \mathrm{~A}$ Resistance of the wire $(\mathrm{R})=25 \Omega$ Time $=1 \mathrm{~min}=60 \mathrm{sec}$ We know that, Joule's heating effect, heat developed across the wire, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\mathrm{H}=(2)^{2} \times 25 \times 60$ $\mathrm{H}=4 \times 25 \times 60$ $\mathrm{H}=6000 \mathrm{~J}$ $\mathrm{H}=6 \times 10^{3} \mathrm{~J}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152649
The power dissipated by a resistance $R$, carrying a sinusoidal $\mathrm{AC}$ current of peak value $I_{P}$ is
1 $I_{P}^{2} R \cos \theta$
2 $\frac{1}{2} I_{P}^{2} R$
3 $\frac{4}{\pi} I_{P}^{2} R$
4 $\frac{1}{\pi} I_{P}^{2} R$
Explanation:
B $\text { Resistance }=\mathrm{R}$ Peak current $\left(\mathrm{I}_{\mathrm{P}}\right)=\sqrt{2} \mathrm{I}_{\mathrm{rms}}$ Peak voltage $\left(\mathrm{V}_{\mathrm{P}}\right)=\sqrt{2} \mathrm{~V}_{\mathrm{rms}}$ Power $=V_{\text {rms }} \times I_{\text {rms }}$ $=\frac{\mathrm{V}_{\mathrm{P}}}{\sqrt{2}} \times \frac{\mathrm{I}_{\mathrm{P}}}{\sqrt{2}}$ $=\frac{1}{2}\left(\mathrm{I}_{\mathrm{P}}\right)\left(\mathrm{I}_{\mathrm{P}}\right) \mathrm{R} \quad\left[\because \mathrm{V}_{\mathrm{P}}=\mathrm{I}_{\mathrm{P}} \mathrm{R}\right]$ $\mathrm{P}=\frac{1}{2} \mathrm{I}_{\mathrm{P}}^{2} \mathrm{R}$
AP EAMCET-25.09.2020
Current Electricity
152650
Two resistances $A(24 \Omega)$ and $B(6 \Omega)$ are connected in series to a d.c. supply. The ratio of heat generated $W_{A}: W_{B}$ Will be
1 $1: 16$
2 $16: 1$
3 $4: 1$
4 $1: 4$
Explanation:
C We know that, Heat generation $=I^{2} R t$ For resistance $\mathrm{A}$, Heat generation in $\mathrm{A}\left(\mathrm{W}_{\mathrm{A}}\right)=\mathrm{I}_{1}{ }^{2} \times 24 \times \mathrm{t}$ For resistance B, Heat generation in $\mathrm{B}\left(\mathrm{W}_{\mathrm{B}}\right)=\mathrm{I}_{2}{ }^{2} \times 6 \times \mathrm{t}$ $\mathrm{I}_{1}=\mathrm{I}_{2}$ because both are in series. The ratios of $\mathrm{W}_{\mathrm{A}}$ and $\mathrm{W}_{\mathrm{B}}$ is- $\frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{\mathrm{I}_{1}^{2} \times 24 \times \mathrm{t}}{\mathrm{I}_{2}^{2} \times 6 \times \mathrm{t}}$ $\frac{\mathrm{W}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{B}}}=\frac{4}{1}$ $\mathrm{~W}_{\mathrm{A}}: \mathrm{W}_{\mathrm{B}}=4: 1$
Tripura-2020
Current Electricity
152651
A bulb of $100 \mathrm{~W}$ rating is connected with $220 \mathrm{~V}$ supply. The resistance of bulb is
1 $484 \Omega$
2 $484{\Omega \mathrm{m}^{-1}}$
3 $2.2 \Omega$
4 $2.2 \times 10^{-3} \Omega \mathrm{m}^{-1}$
Explanation:
A Given, $\mathrm{P}=100 \mathrm{~W}$ $\mathrm{~V}=220 \mathrm{~V}$ The resistance of the bulb, $\mathrm{R} =\frac{\mathrm{V}^{2}}{\mathrm{P}}$ $=\frac{220 \times 220}{100}=484 \Omega$
GUJCET 18.04.2022
Current Electricity
152652
An electric current of 2 A passes through a wire of resistance $25 \Omega$. How much heat will be generated in $1 \mathrm{~min}$ ?
1 $6 \times 10^{3} \mathrm{~J}$
2 $3.6 \times 10^{3} \mathrm{~J}$
3 $0.6 \times 10^{3} \mathrm{~J}$
4 $0.36 \times 10^{3} \mathrm{~J}$
Explanation:
A Given, Current through the wire (I) $=2 \mathrm{~A}$ Resistance of the wire $(\mathrm{R})=25 \Omega$ Time $=1 \mathrm{~min}=60 \mathrm{sec}$ We know that, Joule's heating effect, heat developed across the wire, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}$ $\mathrm{H}=(2)^{2} \times 25 \times 60$ $\mathrm{H}=4 \times 25 \times 60$ $\mathrm{H}=6000 \mathrm{~J}$ $\mathrm{H}=6 \times 10^{3} \mathrm{~J}$