05. Heating Effect of Current (Energy, Power)
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Current Electricity

152671 In the circuit shown in figure, power developed across $1 \Omega, 2 \Omega$ and $3 \Omega$ resistances are in the ratio

1 $1: 2: 3$
2 $4: 2: 27$
3 $6: 4: 9$
4 $2: 127$
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152672 The bulb which glows with maximum intensity in the given circuit is

1 $4 \Omega$ bulb
2 $2 \Omega$ bulb
3 $3 \Omega$ bulb
4 $6 \Omega$ bulb
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152674 A $500 \Omega$ resistor connected to an external battery is placed inside a thermally insulated cylinder fitted with a frictionless piston. The cylinder contains an ideal gas. A current $i$ of $200 \mathrm{~mA}$ flows through the resistor as shown in the figure. The mass of the piston is $10 \mathrm{~kg}$.
Assuming $g=10 \mathrm{~m} / \mathrm{s}^{2}$, the speed at which the piston will move upward, due to heat dissipated by the resistor, so that the temperature of the gas remains unchanged is

1 $10 \mathrm{~cm} / \mathrm{s}$
2 $15 \mathrm{~cm} / \mathrm{s}$
3 $20 \mathrm{~cm} / \mathrm{s}$
4 $30 \mathrm{~cm} / \mathrm{s}$
TS- EAMCET-05.05.2018
Current Electricity

152675 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1 $20 \Omega$
2 $15 \Omega$
3 $10 \Omega$
4 Ans: c
Exp:C Given,
Power dissipated $(\mathrm{P})=30 \mathrm{~W}$
We know that,
Power dissipated $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
Now, resistance $\mathrm{R}$ and $5 \Omega$ are in parallel,
$\therefore \quad$ Equivalent $=\mathrm{R}_{\mathrm{eq}}$
$\therefore \quad \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{5}$
$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{5+\mathrm{R}}{5 \mathrm{R}}$
$\therefore \quad \mathrm{R}_{\text {eq }}=\frac{5 \mathrm{R}}{\mathrm{R}+5}$
Now, $\quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
$30=\frac{10^{2}}{\left(\frac{5 R}{R+5}\right)}$
$30=\frac{100}{\frac{5 R}{R+5}}$
$\frac{5 \mathrm{R}}{\mathrm{R}+5}=\frac{10}{3}$
$3 \mathrm{R}=2 \mathrm{R}+10$
$\mathrm{R}=10 \Omega$
Current Electricity

152676 In the circuit shown, the heat produced in $5 \Omega$ resistor is $10 \mathrm{cal} / \mathrm{s}$. The heat produced per second in $4 \Omega$ resistor will be

1 $1 \mathrm{cal}$
2 $2 \mathrm{cal}$
3 $3 \mathrm{cal}$
4 $4 \mathrm{cal}$
NEET DIGITAL LIBRARY
Current Electricity

152671 In the circuit shown in figure, power developed across $1 \Omega, 2 \Omega$ and $3 \Omega$ resistances are in the ratio

1 $1: 2: 3$
2 $4: 2: 27$
3 $6: 4: 9$
4 $2: 127$
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152672 The bulb which glows with maximum intensity in the given circuit is

1 $4 \Omega$ bulb
2 $2 \Omega$ bulb
3 $3 \Omega$ bulb
4 $6 \Omega$ bulb
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152674 A $500 \Omega$ resistor connected to an external battery is placed inside a thermally insulated cylinder fitted with a frictionless piston. The cylinder contains an ideal gas. A current $i$ of $200 \mathrm{~mA}$ flows through the resistor as shown in the figure. The mass of the piston is $10 \mathrm{~kg}$.
Assuming $g=10 \mathrm{~m} / \mathrm{s}^{2}$, the speed at which the piston will move upward, due to heat dissipated by the resistor, so that the temperature of the gas remains unchanged is

1 $10 \mathrm{~cm} / \mathrm{s}$
2 $15 \mathrm{~cm} / \mathrm{s}$
3 $20 \mathrm{~cm} / \mathrm{s}$
4 $30 \mathrm{~cm} / \mathrm{s}$
TS- EAMCET-05.05.2018
Current Electricity

152675 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1 $20 \Omega$
2 $15 \Omega$
3 $10 \Omega$
4 Ans: c
Exp:C Given,
Power dissipated $(\mathrm{P})=30 \mathrm{~W}$
We know that,
Power dissipated $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
Now, resistance $\mathrm{R}$ and $5 \Omega$ are in parallel,
$\therefore \quad$ Equivalent $=\mathrm{R}_{\mathrm{eq}}$
$\therefore \quad \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{5}$
$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{5+\mathrm{R}}{5 \mathrm{R}}$
$\therefore \quad \mathrm{R}_{\text {eq }}=\frac{5 \mathrm{R}}{\mathrm{R}+5}$
Now, $\quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
$30=\frac{10^{2}}{\left(\frac{5 R}{R+5}\right)}$
$30=\frac{100}{\frac{5 R}{R+5}}$
$\frac{5 \mathrm{R}}{\mathrm{R}+5}=\frac{10}{3}$
$3 \mathrm{R}=2 \mathrm{R}+10$
$\mathrm{R}=10 \Omega$
Current Electricity

152676 In the circuit shown, the heat produced in $5 \Omega$ resistor is $10 \mathrm{cal} / \mathrm{s}$. The heat produced per second in $4 \Omega$ resistor will be

1 $1 \mathrm{cal}$
2 $2 \mathrm{cal}$
3 $3 \mathrm{cal}$
4 $4 \mathrm{cal}$
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Current Electricity

152671 In the circuit shown in figure, power developed across $1 \Omega, 2 \Omega$ and $3 \Omega$ resistances are in the ratio

1 $1: 2: 3$
2 $4: 2: 27$
3 $6: 4: 9$
4 $2: 127$
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152672 The bulb which glows with maximum intensity in the given circuit is

1 $4 \Omega$ bulb
2 $2 \Omega$ bulb
3 $3 \Omega$ bulb
4 $6 \Omega$ bulb
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152674 A $500 \Omega$ resistor connected to an external battery is placed inside a thermally insulated cylinder fitted with a frictionless piston. The cylinder contains an ideal gas. A current $i$ of $200 \mathrm{~mA}$ flows through the resistor as shown in the figure. The mass of the piston is $10 \mathrm{~kg}$.
Assuming $g=10 \mathrm{~m} / \mathrm{s}^{2}$, the speed at which the piston will move upward, due to heat dissipated by the resistor, so that the temperature of the gas remains unchanged is

1 $10 \mathrm{~cm} / \mathrm{s}$
2 $15 \mathrm{~cm} / \mathrm{s}$
3 $20 \mathrm{~cm} / \mathrm{s}$
4 $30 \mathrm{~cm} / \mathrm{s}$
TS- EAMCET-05.05.2018
Current Electricity

152675 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1 $20 \Omega$
2 $15 \Omega$
3 $10 \Omega$
4 Ans: c
Exp:C Given,
Power dissipated $(\mathrm{P})=30 \mathrm{~W}$
We know that,
Power dissipated $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
Now, resistance $\mathrm{R}$ and $5 \Omega$ are in parallel,
$\therefore \quad$ Equivalent $=\mathrm{R}_{\mathrm{eq}}$
$\therefore \quad \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{5}$
$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{5+\mathrm{R}}{5 \mathrm{R}}$
$\therefore \quad \mathrm{R}_{\text {eq }}=\frac{5 \mathrm{R}}{\mathrm{R}+5}$
Now, $\quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
$30=\frac{10^{2}}{\left(\frac{5 R}{R+5}\right)}$
$30=\frac{100}{\frac{5 R}{R+5}}$
$\frac{5 \mathrm{R}}{\mathrm{R}+5}=\frac{10}{3}$
$3 \mathrm{R}=2 \mathrm{R}+10$
$\mathrm{R}=10 \Omega$
Current Electricity

152676 In the circuit shown, the heat produced in $5 \Omega$ resistor is $10 \mathrm{cal} / \mathrm{s}$. The heat produced per second in $4 \Omega$ resistor will be

1 $1 \mathrm{cal}$
2 $2 \mathrm{cal}$
3 $3 \mathrm{cal}$
4 $4 \mathrm{cal}$
For More Updates join with Us
Current Electricity

152671 In the circuit shown in figure, power developed across $1 \Omega, 2 \Omega$ and $3 \Omega$ resistances are in the ratio

1 $1: 2: 3$
2 $4: 2: 27$
3 $6: 4: 9$
4 $2: 127$
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152672 The bulb which glows with maximum intensity in the given circuit is

1 $4 \Omega$ bulb
2 $2 \Omega$ bulb
3 $3 \Omega$ bulb
4 $6 \Omega$ bulb
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152674 A $500 \Omega$ resistor connected to an external battery is placed inside a thermally insulated cylinder fitted with a frictionless piston. The cylinder contains an ideal gas. A current $i$ of $200 \mathrm{~mA}$ flows through the resistor as shown in the figure. The mass of the piston is $10 \mathrm{~kg}$.
Assuming $g=10 \mathrm{~m} / \mathrm{s}^{2}$, the speed at which the piston will move upward, due to heat dissipated by the resistor, so that the temperature of the gas remains unchanged is

1 $10 \mathrm{~cm} / \mathrm{s}$
2 $15 \mathrm{~cm} / \mathrm{s}$
3 $20 \mathrm{~cm} / \mathrm{s}$
4 $30 \mathrm{~cm} / \mathrm{s}$
TS- EAMCET-05.05.2018
Current Electricity

152675 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1 $20 \Omega$
2 $15 \Omega$
3 $10 \Omega$
4 Ans: c
Exp:C Given,
Power dissipated $(\mathrm{P})=30 \mathrm{~W}$
We know that,
Power dissipated $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
Now, resistance $\mathrm{R}$ and $5 \Omega$ are in parallel,
$\therefore \quad$ Equivalent $=\mathrm{R}_{\mathrm{eq}}$
$\therefore \quad \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{5}$
$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{5+\mathrm{R}}{5 \mathrm{R}}$
$\therefore \quad \mathrm{R}_{\text {eq }}=\frac{5 \mathrm{R}}{\mathrm{R}+5}$
Now, $\quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
$30=\frac{10^{2}}{\left(\frac{5 R}{R+5}\right)}$
$30=\frac{100}{\frac{5 R}{R+5}}$
$\frac{5 \mathrm{R}}{\mathrm{R}+5}=\frac{10}{3}$
$3 \mathrm{R}=2 \mathrm{R}+10$
$\mathrm{R}=10 \Omega$
Current Electricity

152676 In the circuit shown, the heat produced in $5 \Omega$ resistor is $10 \mathrm{cal} / \mathrm{s}$. The heat produced per second in $4 \Omega$ resistor will be

1 $1 \mathrm{cal}$
2 $2 \mathrm{cal}$
3 $3 \mathrm{cal}$
4 $4 \mathrm{cal}$
NEET DIGITAL LIBRARY
Current Electricity

152671 In the circuit shown in figure, power developed across $1 \Omega, 2 \Omega$ and $3 \Omega$ resistances are in the ratio

1 $1: 2: 3$
2 $4: 2: 27$
3 $6: 4: 9$
4 $2: 127$
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152672 The bulb which glows with maximum intensity in the given circuit is

1 $4 \Omega$ bulb
2 $2 \Omega$ bulb
3 $3 \Omega$ bulb
4 $6 \Omega$ bulb
AP EAMCET (23.04.2018) Shift-2
Current Electricity

152674 A $500 \Omega$ resistor connected to an external battery is placed inside a thermally insulated cylinder fitted with a frictionless piston. The cylinder contains an ideal gas. A current $i$ of $200 \mathrm{~mA}$ flows through the resistor as shown in the figure. The mass of the piston is $10 \mathrm{~kg}$.
Assuming $g=10 \mathrm{~m} / \mathrm{s}^{2}$, the speed at which the piston will move upward, due to heat dissipated by the resistor, so that the temperature of the gas remains unchanged is

1 $10 \mathrm{~cm} / \mathrm{s}$
2 $15 \mathrm{~cm} / \mathrm{s}$
3 $20 \mathrm{~cm} / \mathrm{s}$
4 $30 \mathrm{~cm} / \mathrm{s}$
TS- EAMCET-05.05.2018
Current Electricity

152675 The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

1 $20 \Omega$
2 $15 \Omega$
3 $10 \Omega$
4 Ans: c
Exp:C Given,
Power dissipated $(\mathrm{P})=30 \mathrm{~W}$
We know that,
Power dissipated $(\mathrm{P})=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
Now, resistance $\mathrm{R}$ and $5 \Omega$ are in parallel,
$\therefore \quad$ Equivalent $=\mathrm{R}_{\mathrm{eq}}$
$\therefore \quad \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{5}$
$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{5+\mathrm{R}}{5 \mathrm{R}}$
$\therefore \quad \mathrm{R}_{\text {eq }}=\frac{5 \mathrm{R}}{\mathrm{R}+5}$
Now, $\quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}_{\mathrm{eq}}}$
$30=\frac{10^{2}}{\left(\frac{5 R}{R+5}\right)}$
$30=\frac{100}{\frac{5 R}{R+5}}$
$\frac{5 \mathrm{R}}{\mathrm{R}+5}=\frac{10}{3}$
$3 \mathrm{R}=2 \mathrm{R}+10$
$\mathrm{R}=10 \Omega$
Current Electricity

152676 In the circuit shown, the heat produced in $5 \Omega$ resistor is $10 \mathrm{cal} / \mathrm{s}$. The heat produced per second in $4 \Omega$ resistor will be

1 $1 \mathrm{cal}$
2 $2 \mathrm{cal}$
3 $3 \mathrm{cal}$
4 $4 \mathrm{cal}$