04. Cells, Internal Resistance and Cell Combination, Thermocouple
Current Electricity

152563 In a thermocouple, the neutral temperature is $270^{\circ} \mathrm{C}$ and the temperature of inversion is $525^{\circ} \mathrm{C}$. The temperature of cold junction would be

1 $30^{\circ} \mathrm{C}$
2 $255^{\circ} \mathrm{C}$
3 $15^{\circ} \mathrm{C}$
4 $25^{\circ} \mathrm{C}$
Current Electricity

152565 Two diode having resistance $20 \Omega$ and is centre tapped with potential difference $50 \mathrm{~V}$. If external resistance is $980 \Omega$, what is current through resistance?

1 $0.05 \mathrm{~A}$
2 $0.025 \mathrm{~A}$
3 $0.25 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Current Electricity

152566 The range of voltmeter is $10 \mathrm{~V}$ and its internal resistance is $50 \Omega$. To convert it to a voltmeter of range $15 \mathrm{~V}$, how much resistance is to be added?

1 Add $25 \Omega$ resistor in parallel
2 Add $25 \Omega$ resistor in series
3 Add $125 \Omega$ resistor in parallel
4 Add $125 \Omega$ resistor in series
Current Electricity

152567 The emf of a cell $E$ is $15 \mathrm{~V}$ as shown in the figure with an internal resistance of $0.5 \Omega$. Then the value of the current drawn from the cell is

1 $3 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $1 \mathrm{~A}$
Current Electricity

152563 In a thermocouple, the neutral temperature is $270^{\circ} \mathrm{C}$ and the temperature of inversion is $525^{\circ} \mathrm{C}$. The temperature of cold junction would be

1 $30^{\circ} \mathrm{C}$
2 $255^{\circ} \mathrm{C}$
3 $15^{\circ} \mathrm{C}$
4 $25^{\circ} \mathrm{C}$
Current Electricity

152565 Two diode having resistance $20 \Omega$ and is centre tapped with potential difference $50 \mathrm{~V}$. If external resistance is $980 \Omega$, what is current through resistance?

1 $0.05 \mathrm{~A}$
2 $0.025 \mathrm{~A}$
3 $0.25 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Current Electricity

152566 The range of voltmeter is $10 \mathrm{~V}$ and its internal resistance is $50 \Omega$. To convert it to a voltmeter of range $15 \mathrm{~V}$, how much resistance is to be added?

1 Add $25 \Omega$ resistor in parallel
2 Add $25 \Omega$ resistor in series
3 Add $125 \Omega$ resistor in parallel
4 Add $125 \Omega$ resistor in series
Current Electricity

152567 The emf of a cell $E$ is $15 \mathrm{~V}$ as shown in the figure with an internal resistance of $0.5 \Omega$. Then the value of the current drawn from the cell is

1 $3 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $1 \mathrm{~A}$
Current Electricity

152563 In a thermocouple, the neutral temperature is $270^{\circ} \mathrm{C}$ and the temperature of inversion is $525^{\circ} \mathrm{C}$. The temperature of cold junction would be

1 $30^{\circ} \mathrm{C}$
2 $255^{\circ} \mathrm{C}$
3 $15^{\circ} \mathrm{C}$
4 $25^{\circ} \mathrm{C}$
Current Electricity

152565 Two diode having resistance $20 \Omega$ and is centre tapped with potential difference $50 \mathrm{~V}$. If external resistance is $980 \Omega$, what is current through resistance?

1 $0.05 \mathrm{~A}$
2 $0.025 \mathrm{~A}$
3 $0.25 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Current Electricity

152566 The range of voltmeter is $10 \mathrm{~V}$ and its internal resistance is $50 \Omega$. To convert it to a voltmeter of range $15 \mathrm{~V}$, how much resistance is to be added?

1 Add $25 \Omega$ resistor in parallel
2 Add $25 \Omega$ resistor in series
3 Add $125 \Omega$ resistor in parallel
4 Add $125 \Omega$ resistor in series
Current Electricity

152567 The emf of a cell $E$ is $15 \mathrm{~V}$ as shown in the figure with an internal resistance of $0.5 \Omega$. Then the value of the current drawn from the cell is

1 $3 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $1 \mathrm{~A}$
Current Electricity

152563 In a thermocouple, the neutral temperature is $270^{\circ} \mathrm{C}$ and the temperature of inversion is $525^{\circ} \mathrm{C}$. The temperature of cold junction would be

1 $30^{\circ} \mathrm{C}$
2 $255^{\circ} \mathrm{C}$
3 $15^{\circ} \mathrm{C}$
4 $25^{\circ} \mathrm{C}$
Current Electricity

152565 Two diode having resistance $20 \Omega$ and is centre tapped with potential difference $50 \mathrm{~V}$. If external resistance is $980 \Omega$, what is current through resistance?

1 $0.05 \mathrm{~A}$
2 $0.025 \mathrm{~A}$
3 $0.25 \mathrm{~A}$
4 $0.5 \mathrm{~A}$
Current Electricity

152566 The range of voltmeter is $10 \mathrm{~V}$ and its internal resistance is $50 \Omega$. To convert it to a voltmeter of range $15 \mathrm{~V}$, how much resistance is to be added?

1 Add $25 \Omega$ resistor in parallel
2 Add $25 \Omega$ resistor in series
3 Add $125 \Omega$ resistor in parallel
4 Add $125 \Omega$ resistor in series
Current Electricity

152567 The emf of a cell $E$ is $15 \mathrm{~V}$ as shown in the figure with an internal resistance of $0.5 \Omega$. Then the value of the current drawn from the cell is

1 $3 \mathrm{~A}$
2 $2 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $1 \mathrm{~A}$