152667
The resistor $R_{1}=3 \Omega$ and $R_{2}=1 \Omega$ are connected in parallel to a $20 \mathrm{~V}$ battery. Find the heat developed in the resistor $R_{1}$ in one minute
1 $600 \mathrm{~J}$
2 $800 \mathrm{~J}$
3 $6000 \mathrm{~J}$
4 $8000 \mathrm{~J}$
5 $7000 \mathrm{~J}$
Explanation:
D Given, Resistance $\left(\mathrm{R}_{1}\right)=3 \Omega$ Resistance $\left(\mathrm{R}_{2}\right)=1 \Omega$ Voltage $(\mathrm{V})=20 \mathrm{~V}$ We know that, Heat developed $(\mathrm{Q})=\frac{\mathrm{V}^{2}}{\mathrm{R}} \times \mathrm{t}$ $=I^{2} \mathrm{Rt}$ The connection is parallel, so each branch of the circuit has potential drop of $20 \mathrm{~V}$. $\therefore$ Current in $\mathrm{R}_{1}$ resistor, $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{1}}=\frac{20}{3} \mathrm{~A}$ $\therefore \quad \mathrm{Q}=\mathrm{I}^{2} \mathrm{R}_{1} \times \mathrm{t}$ $=\left(\frac{20}{3}\right)^{2} \times 3 \times 60$ $=\frac{20 \times 20 \times 3 \times 60}{3 \times 3}$ $\mathrm{Q}=8000 \mathrm{~J}$
Kerala CEE-2019
Current Electricity
152668
An electric bulb rated $100 \mathrm{~W}$ at $220 \mathrm{~V}$ is operating at $110 \mathrm{~V}$. What is the power consumed?
1 $50 \mathrm{~W}$
2 $75 \mathrm{~W}$
3 $100 \mathrm{~W}$
4 $25 \mathrm{~W}$
Explanation:
D Given, Rated power, $\mathrm{P}=100 \mathrm{~W}$ Voltage, $\mathrm{V}=220 \mathrm{Volt}$ As we know, Power $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore$ Resistance $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now, Power at 110 voltage, $=\frac{220^{2}}{100}=\frac{48400}{100}$ $=484 \mathrm{Ohm}$ $\therefore \quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{110^{2}}{484}$ $\mathrm{P}=25 \mathrm{~W}$
J and K CET- 2019
Current Electricity
152669
Two bulbs of $500 \mathrm{~W}$ and $200 \mathrm{~W}$ are manufactured to operate on $220 \mathrm{~V}$ line. The ratio of heat produced in $500 \mathrm{~W}$ and $200 \mathrm{~W}$, in two cases, when firstly they are connected in parallel and secondary in series will be
1 $\frac{5}{2}: \frac{2}{5}$
2 $\frac{5}{2}: \frac{5}{2}$
3 $\frac{2}{5}: \frac{5}{2}$
4 $\frac{2}{5}: \frac{2}{5}$
Explanation:
A Given, $\mathrm{P}_{1}=500 \mathrm{~W}$ $\mathrm{P}_{2}=200 \mathrm{~W}$ Firstly, two bulbs connected in parallel. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{P}_{1} \mathrm{t}}{\mathrm{P}_{2} \mathrm{t}}=\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{500}{200}=\frac{5}{2}$ Secondary, two bulbs connected in series. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{I}^{2} \mathrm{R}_{1} \mathrm{t}}{\mathrm{I}^{2} \mathrm{R}_{2} \mathrm{t}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{V}^{2}}{\mathrm{P}^{1}}}{\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{200}{500}=\frac{2}{5}$ Therefore, the ratio of heat generated in parallel to series is $\frac{5}{2}: \frac{2}{5}$.
AIIMS-26.05.2018(E)
Current Electricity
152670
A resistor develops $800 \mathrm{~J}$ of thermal energy is $20 \mathrm{~s}$ when a current of $4 \mathrm{~A}$ is passed through it. If the current is increased to $8 \mathrm{~A}$, the energy developed in $20 \mathrm{~s}$ is
1 $800 \mathrm{~J}$
2 $1600 \mathrm{~J}$
3 $3200 \mathrm{~J}$
4 $400 \mathrm{~J}$
Explanation:
C Given, $\mathrm{Q}_{1}=800 \mathrm{~J}, \mathrm{Q}_{2}=?$ $\mathrm{t}_{1}=20 \mathrm{sec}, \mathrm{t}_{2}=20 \mathrm{sec}$ $\mathrm{i}_{1}=4 \mathrm{~A}, \mathrm{i}_{2}=8 \mathrm{~A}$ $\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}$ We know that, $\mathrm{Q}_{1}=\mathrm{i}_{1}^{2} \mathrm{Rt}_{1}$ $800=(4)^{2} \times \mathrm{R} \times 20$ $\mathrm{R}=\frac{800}{320}=\frac{5}{2} \Omega$ The thermal energy developed $\left(\mathrm{Q}_{2}\right)$, when the current is $8 \mathrm{~A}$, is $\mathrm{Q}_{2}=\mathrm{i}_{2}^{2} \mathrm{Rt}_{2}$ $\mathrm{Q}_{2}=(8)^{2} \times \frac{5}{2} \times 20$ $\mathrm{Q}_{2}=64 \times 50$ $\mathrm{Q}_{2}=3200 \mathrm{~J}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152667
The resistor $R_{1}=3 \Omega$ and $R_{2}=1 \Omega$ are connected in parallel to a $20 \mathrm{~V}$ battery. Find the heat developed in the resistor $R_{1}$ in one minute
1 $600 \mathrm{~J}$
2 $800 \mathrm{~J}$
3 $6000 \mathrm{~J}$
4 $8000 \mathrm{~J}$
5 $7000 \mathrm{~J}$
Explanation:
D Given, Resistance $\left(\mathrm{R}_{1}\right)=3 \Omega$ Resistance $\left(\mathrm{R}_{2}\right)=1 \Omega$ Voltage $(\mathrm{V})=20 \mathrm{~V}$ We know that, Heat developed $(\mathrm{Q})=\frac{\mathrm{V}^{2}}{\mathrm{R}} \times \mathrm{t}$ $=I^{2} \mathrm{Rt}$ The connection is parallel, so each branch of the circuit has potential drop of $20 \mathrm{~V}$. $\therefore$ Current in $\mathrm{R}_{1}$ resistor, $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{1}}=\frac{20}{3} \mathrm{~A}$ $\therefore \quad \mathrm{Q}=\mathrm{I}^{2} \mathrm{R}_{1} \times \mathrm{t}$ $=\left(\frac{20}{3}\right)^{2} \times 3 \times 60$ $=\frac{20 \times 20 \times 3 \times 60}{3 \times 3}$ $\mathrm{Q}=8000 \mathrm{~J}$
Kerala CEE-2019
Current Electricity
152668
An electric bulb rated $100 \mathrm{~W}$ at $220 \mathrm{~V}$ is operating at $110 \mathrm{~V}$. What is the power consumed?
1 $50 \mathrm{~W}$
2 $75 \mathrm{~W}$
3 $100 \mathrm{~W}$
4 $25 \mathrm{~W}$
Explanation:
D Given, Rated power, $\mathrm{P}=100 \mathrm{~W}$ Voltage, $\mathrm{V}=220 \mathrm{Volt}$ As we know, Power $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore$ Resistance $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now, Power at 110 voltage, $=\frac{220^{2}}{100}=\frac{48400}{100}$ $=484 \mathrm{Ohm}$ $\therefore \quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{110^{2}}{484}$ $\mathrm{P}=25 \mathrm{~W}$
J and K CET- 2019
Current Electricity
152669
Two bulbs of $500 \mathrm{~W}$ and $200 \mathrm{~W}$ are manufactured to operate on $220 \mathrm{~V}$ line. The ratio of heat produced in $500 \mathrm{~W}$ and $200 \mathrm{~W}$, in two cases, when firstly they are connected in parallel and secondary in series will be
1 $\frac{5}{2}: \frac{2}{5}$
2 $\frac{5}{2}: \frac{5}{2}$
3 $\frac{2}{5}: \frac{5}{2}$
4 $\frac{2}{5}: \frac{2}{5}$
Explanation:
A Given, $\mathrm{P}_{1}=500 \mathrm{~W}$ $\mathrm{P}_{2}=200 \mathrm{~W}$ Firstly, two bulbs connected in parallel. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{P}_{1} \mathrm{t}}{\mathrm{P}_{2} \mathrm{t}}=\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{500}{200}=\frac{5}{2}$ Secondary, two bulbs connected in series. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{I}^{2} \mathrm{R}_{1} \mathrm{t}}{\mathrm{I}^{2} \mathrm{R}_{2} \mathrm{t}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{V}^{2}}{\mathrm{P}^{1}}}{\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{200}{500}=\frac{2}{5}$ Therefore, the ratio of heat generated in parallel to series is $\frac{5}{2}: \frac{2}{5}$.
AIIMS-26.05.2018(E)
Current Electricity
152670
A resistor develops $800 \mathrm{~J}$ of thermal energy is $20 \mathrm{~s}$ when a current of $4 \mathrm{~A}$ is passed through it. If the current is increased to $8 \mathrm{~A}$, the energy developed in $20 \mathrm{~s}$ is
1 $800 \mathrm{~J}$
2 $1600 \mathrm{~J}$
3 $3200 \mathrm{~J}$
4 $400 \mathrm{~J}$
Explanation:
C Given, $\mathrm{Q}_{1}=800 \mathrm{~J}, \mathrm{Q}_{2}=?$ $\mathrm{t}_{1}=20 \mathrm{sec}, \mathrm{t}_{2}=20 \mathrm{sec}$ $\mathrm{i}_{1}=4 \mathrm{~A}, \mathrm{i}_{2}=8 \mathrm{~A}$ $\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}$ We know that, $\mathrm{Q}_{1}=\mathrm{i}_{1}^{2} \mathrm{Rt}_{1}$ $800=(4)^{2} \times \mathrm{R} \times 20$ $\mathrm{R}=\frac{800}{320}=\frac{5}{2} \Omega$ The thermal energy developed $\left(\mathrm{Q}_{2}\right)$, when the current is $8 \mathrm{~A}$, is $\mathrm{Q}_{2}=\mathrm{i}_{2}^{2} \mathrm{Rt}_{2}$ $\mathrm{Q}_{2}=(8)^{2} \times \frac{5}{2} \times 20$ $\mathrm{Q}_{2}=64 \times 50$ $\mathrm{Q}_{2}=3200 \mathrm{~J}$
152667
The resistor $R_{1}=3 \Omega$ and $R_{2}=1 \Omega$ are connected in parallel to a $20 \mathrm{~V}$ battery. Find the heat developed in the resistor $R_{1}$ in one minute
1 $600 \mathrm{~J}$
2 $800 \mathrm{~J}$
3 $6000 \mathrm{~J}$
4 $8000 \mathrm{~J}$
5 $7000 \mathrm{~J}$
Explanation:
D Given, Resistance $\left(\mathrm{R}_{1}\right)=3 \Omega$ Resistance $\left(\mathrm{R}_{2}\right)=1 \Omega$ Voltage $(\mathrm{V})=20 \mathrm{~V}$ We know that, Heat developed $(\mathrm{Q})=\frac{\mathrm{V}^{2}}{\mathrm{R}} \times \mathrm{t}$ $=I^{2} \mathrm{Rt}$ The connection is parallel, so each branch of the circuit has potential drop of $20 \mathrm{~V}$. $\therefore$ Current in $\mathrm{R}_{1}$ resistor, $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{1}}=\frac{20}{3} \mathrm{~A}$ $\therefore \quad \mathrm{Q}=\mathrm{I}^{2} \mathrm{R}_{1} \times \mathrm{t}$ $=\left(\frac{20}{3}\right)^{2} \times 3 \times 60$ $=\frac{20 \times 20 \times 3 \times 60}{3 \times 3}$ $\mathrm{Q}=8000 \mathrm{~J}$
Kerala CEE-2019
Current Electricity
152668
An electric bulb rated $100 \mathrm{~W}$ at $220 \mathrm{~V}$ is operating at $110 \mathrm{~V}$. What is the power consumed?
1 $50 \mathrm{~W}$
2 $75 \mathrm{~W}$
3 $100 \mathrm{~W}$
4 $25 \mathrm{~W}$
Explanation:
D Given, Rated power, $\mathrm{P}=100 \mathrm{~W}$ Voltage, $\mathrm{V}=220 \mathrm{Volt}$ As we know, Power $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore$ Resistance $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now, Power at 110 voltage, $=\frac{220^{2}}{100}=\frac{48400}{100}$ $=484 \mathrm{Ohm}$ $\therefore \quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{110^{2}}{484}$ $\mathrm{P}=25 \mathrm{~W}$
J and K CET- 2019
Current Electricity
152669
Two bulbs of $500 \mathrm{~W}$ and $200 \mathrm{~W}$ are manufactured to operate on $220 \mathrm{~V}$ line. The ratio of heat produced in $500 \mathrm{~W}$ and $200 \mathrm{~W}$, in two cases, when firstly they are connected in parallel and secondary in series will be
1 $\frac{5}{2}: \frac{2}{5}$
2 $\frac{5}{2}: \frac{5}{2}$
3 $\frac{2}{5}: \frac{5}{2}$
4 $\frac{2}{5}: \frac{2}{5}$
Explanation:
A Given, $\mathrm{P}_{1}=500 \mathrm{~W}$ $\mathrm{P}_{2}=200 \mathrm{~W}$ Firstly, two bulbs connected in parallel. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{P}_{1} \mathrm{t}}{\mathrm{P}_{2} \mathrm{t}}=\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{500}{200}=\frac{5}{2}$ Secondary, two bulbs connected in series. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{I}^{2} \mathrm{R}_{1} \mathrm{t}}{\mathrm{I}^{2} \mathrm{R}_{2} \mathrm{t}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{V}^{2}}{\mathrm{P}^{1}}}{\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{200}{500}=\frac{2}{5}$ Therefore, the ratio of heat generated in parallel to series is $\frac{5}{2}: \frac{2}{5}$.
AIIMS-26.05.2018(E)
Current Electricity
152670
A resistor develops $800 \mathrm{~J}$ of thermal energy is $20 \mathrm{~s}$ when a current of $4 \mathrm{~A}$ is passed through it. If the current is increased to $8 \mathrm{~A}$, the energy developed in $20 \mathrm{~s}$ is
1 $800 \mathrm{~J}$
2 $1600 \mathrm{~J}$
3 $3200 \mathrm{~J}$
4 $400 \mathrm{~J}$
Explanation:
C Given, $\mathrm{Q}_{1}=800 \mathrm{~J}, \mathrm{Q}_{2}=?$ $\mathrm{t}_{1}=20 \mathrm{sec}, \mathrm{t}_{2}=20 \mathrm{sec}$ $\mathrm{i}_{1}=4 \mathrm{~A}, \mathrm{i}_{2}=8 \mathrm{~A}$ $\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}$ We know that, $\mathrm{Q}_{1}=\mathrm{i}_{1}^{2} \mathrm{Rt}_{1}$ $800=(4)^{2} \times \mathrm{R} \times 20$ $\mathrm{R}=\frac{800}{320}=\frac{5}{2} \Omega$ The thermal energy developed $\left(\mathrm{Q}_{2}\right)$, when the current is $8 \mathrm{~A}$, is $\mathrm{Q}_{2}=\mathrm{i}_{2}^{2} \mathrm{Rt}_{2}$ $\mathrm{Q}_{2}=(8)^{2} \times \frac{5}{2} \times 20$ $\mathrm{Q}_{2}=64 \times 50$ $\mathrm{Q}_{2}=3200 \mathrm{~J}$
152667
The resistor $R_{1}=3 \Omega$ and $R_{2}=1 \Omega$ are connected in parallel to a $20 \mathrm{~V}$ battery. Find the heat developed in the resistor $R_{1}$ in one minute
1 $600 \mathrm{~J}$
2 $800 \mathrm{~J}$
3 $6000 \mathrm{~J}$
4 $8000 \mathrm{~J}$
5 $7000 \mathrm{~J}$
Explanation:
D Given, Resistance $\left(\mathrm{R}_{1}\right)=3 \Omega$ Resistance $\left(\mathrm{R}_{2}\right)=1 \Omega$ Voltage $(\mathrm{V})=20 \mathrm{~V}$ We know that, Heat developed $(\mathrm{Q})=\frac{\mathrm{V}^{2}}{\mathrm{R}} \times \mathrm{t}$ $=I^{2} \mathrm{Rt}$ The connection is parallel, so each branch of the circuit has potential drop of $20 \mathrm{~V}$. $\therefore$ Current in $\mathrm{R}_{1}$ resistor, $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}_{1}}=\frac{20}{3} \mathrm{~A}$ $\therefore \quad \mathrm{Q}=\mathrm{I}^{2} \mathrm{R}_{1} \times \mathrm{t}$ $=\left(\frac{20}{3}\right)^{2} \times 3 \times 60$ $=\frac{20 \times 20 \times 3 \times 60}{3 \times 3}$ $\mathrm{Q}=8000 \mathrm{~J}$
Kerala CEE-2019
Current Electricity
152668
An electric bulb rated $100 \mathrm{~W}$ at $220 \mathrm{~V}$ is operating at $110 \mathrm{~V}$. What is the power consumed?
1 $50 \mathrm{~W}$
2 $75 \mathrm{~W}$
3 $100 \mathrm{~W}$
4 $25 \mathrm{~W}$
Explanation:
D Given, Rated power, $\mathrm{P}=100 \mathrm{~W}$ Voltage, $\mathrm{V}=220 \mathrm{Volt}$ As we know, Power $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\therefore$ Resistance $(\mathrm{R})=\frac{\mathrm{V}^{2}}{\mathrm{P}}$ Now, Power at 110 voltage, $=\frac{220^{2}}{100}=\frac{48400}{100}$ $=484 \mathrm{Ohm}$ $\therefore \quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{110^{2}}{484}$ $\mathrm{P}=25 \mathrm{~W}$
J and K CET- 2019
Current Electricity
152669
Two bulbs of $500 \mathrm{~W}$ and $200 \mathrm{~W}$ are manufactured to operate on $220 \mathrm{~V}$ line. The ratio of heat produced in $500 \mathrm{~W}$ and $200 \mathrm{~W}$, in two cases, when firstly they are connected in parallel and secondary in series will be
1 $\frac{5}{2}: \frac{2}{5}$
2 $\frac{5}{2}: \frac{5}{2}$
3 $\frac{2}{5}: \frac{5}{2}$
4 $\frac{2}{5}: \frac{2}{5}$
Explanation:
A Given, $\mathrm{P}_{1}=500 \mathrm{~W}$ $\mathrm{P}_{2}=200 \mathrm{~W}$ Firstly, two bulbs connected in parallel. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{P}_{1} \mathrm{t}}{\mathrm{P}_{2} \mathrm{t}}=\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{500}{200}=\frac{5}{2}$ Secondary, two bulbs connected in series. $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{I}^{2} \mathrm{R}_{1} \mathrm{t}}{\mathrm{I}^{2} \mathrm{R}_{2} \mathrm{t}}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\frac{\mathrm{V}^{2}}{\mathrm{P}^{1}}}{\frac{\mathrm{V}^{2}}{\mathrm{P}_{2}}}$ $\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{200}{500}=\frac{2}{5}$ Therefore, the ratio of heat generated in parallel to series is $\frac{5}{2}: \frac{2}{5}$.
AIIMS-26.05.2018(E)
Current Electricity
152670
A resistor develops $800 \mathrm{~J}$ of thermal energy is $20 \mathrm{~s}$ when a current of $4 \mathrm{~A}$ is passed through it. If the current is increased to $8 \mathrm{~A}$, the energy developed in $20 \mathrm{~s}$ is
1 $800 \mathrm{~J}$
2 $1600 \mathrm{~J}$
3 $3200 \mathrm{~J}$
4 $400 \mathrm{~J}$
Explanation:
C Given, $\mathrm{Q}_{1}=800 \mathrm{~J}, \mathrm{Q}_{2}=?$ $\mathrm{t}_{1}=20 \mathrm{sec}, \mathrm{t}_{2}=20 \mathrm{sec}$ $\mathrm{i}_{1}=4 \mathrm{~A}, \mathrm{i}_{2}=8 \mathrm{~A}$ $\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}$ We know that, $\mathrm{Q}_{1}=\mathrm{i}_{1}^{2} \mathrm{Rt}_{1}$ $800=(4)^{2} \times \mathrm{R} \times 20$ $\mathrm{R}=\frac{800}{320}=\frac{5}{2} \Omega$ The thermal energy developed $\left(\mathrm{Q}_{2}\right)$, when the current is $8 \mathrm{~A}$, is $\mathrm{Q}_{2}=\mathrm{i}_{2}^{2} \mathrm{Rt}_{2}$ $\mathrm{Q}_{2}=(8)^{2} \times \frac{5}{2} \times 20$ $\mathrm{Q}_{2}=64 \times 50$ $\mathrm{Q}_{2}=3200 \mathrm{~J}$
