268423
In the circuit shown below, the cell has an emf of \(10 \mathrm{~V}\) and internal resistance of \(1 \mathrm{ohm}\) the other resistances are shown in the figure. The potential difference \(V_{A}-V_{B}\) is
1 \(\mathrm{CV}\)
2 \(4 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(-2 \mathrm{~V}\)
Explanation:
Apply Ohm's law.
Current Electricity
268424
A uniform wire of resistance \(20 \Omega\) having resistance \(1 \Omega\) / m is bent in the form of circle as shown in fig. If the equivalent resistance between \(M\) and \(N\) is \(1.8 \Omega\), then the length of the shorter section is
1 \(2 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(1.8 \mathrm{~m}\)
4 \(18 \mathrm{~m}\)
Explanation:
Let the resistance of shorter part \(\mathrm{MN}\) bex. Then resistance of longer part is \((20-x) \Omega\) \(R_{e q}=\frac{(20-x) x}{20-x+x}=1.8 \Omega\) Solving we get \(x=2 \Omega\) So length of shorter part \(=2 \mathrm{~m}\)
Current Electricity
268425
If the voltmeter reads \(0.2 \mathrm{~V}\) and the ammeter reads \(0.101 \mathrm{~A}\), the resistance of the voltmeter is(in ohm)
1 500
2 1000
3 200
4 400
Explanation:
\(\left[\frac{2 x}{2+x}\right](0.101)=0.2 \quad\) solve for ' \(x\) '.
Current Electricity
268426
In the en then when both switches \(S_{1}, S_{2}\) are closed or opened. The value of resistance \(R\) is
1 \(200 \Omega\)
2 \(100 \Omega\)
3 \(400 \Omega\)
4 \(-300 \Omega\)
Explanation:
When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) areopened, \(\mathrm{i}=\frac{1.5}{450}\) When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) are closed, \(i=\frac{1.5[100+R]}{400 R+30,000}\)
Current Electricity
268427
In the following diagram ammeter reading is \(4 \mathrm{~A}\), voltmeter reading is \(20 \mathrm{~V}\), the value of \(R\) is
1 \(\lt 5 \Omega\)
2 \(<5 \Omega\)
3 \(=5 \Omega\)
4 \(\leq 5 \Omega\)
Explanation:
\(\mathrm{iR}=20\), here \(\mathrm{i}<4 \mathrm{~A} \Rightarrow \mathrm{R}\lt 5 \Omega\)
268423
In the circuit shown below, the cell has an emf of \(10 \mathrm{~V}\) and internal resistance of \(1 \mathrm{ohm}\) the other resistances are shown in the figure. The potential difference \(V_{A}-V_{B}\) is
1 \(\mathrm{CV}\)
2 \(4 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(-2 \mathrm{~V}\)
Explanation:
Apply Ohm's law.
Current Electricity
268424
A uniform wire of resistance \(20 \Omega\) having resistance \(1 \Omega\) / m is bent in the form of circle as shown in fig. If the equivalent resistance between \(M\) and \(N\) is \(1.8 \Omega\), then the length of the shorter section is
1 \(2 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(1.8 \mathrm{~m}\)
4 \(18 \mathrm{~m}\)
Explanation:
Let the resistance of shorter part \(\mathrm{MN}\) bex. Then resistance of longer part is \((20-x) \Omega\) \(R_{e q}=\frac{(20-x) x}{20-x+x}=1.8 \Omega\) Solving we get \(x=2 \Omega\) So length of shorter part \(=2 \mathrm{~m}\)
Current Electricity
268425
If the voltmeter reads \(0.2 \mathrm{~V}\) and the ammeter reads \(0.101 \mathrm{~A}\), the resistance of the voltmeter is(in ohm)
1 500
2 1000
3 200
4 400
Explanation:
\(\left[\frac{2 x}{2+x}\right](0.101)=0.2 \quad\) solve for ' \(x\) '.
Current Electricity
268426
In the en then when both switches \(S_{1}, S_{2}\) are closed or opened. The value of resistance \(R\) is
1 \(200 \Omega\)
2 \(100 \Omega\)
3 \(400 \Omega\)
4 \(-300 \Omega\)
Explanation:
When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) areopened, \(\mathrm{i}=\frac{1.5}{450}\) When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) are closed, \(i=\frac{1.5[100+R]}{400 R+30,000}\)
Current Electricity
268427
In the following diagram ammeter reading is \(4 \mathrm{~A}\), voltmeter reading is \(20 \mathrm{~V}\), the value of \(R\) is
1 \(\lt 5 \Omega\)
2 \(<5 \Omega\)
3 \(=5 \Omega\)
4 \(\leq 5 \Omega\)
Explanation:
\(\mathrm{iR}=20\), here \(\mathrm{i}<4 \mathrm{~A} \Rightarrow \mathrm{R}\lt 5 \Omega\)
268423
In the circuit shown below, the cell has an emf of \(10 \mathrm{~V}\) and internal resistance of \(1 \mathrm{ohm}\) the other resistances are shown in the figure. The potential difference \(V_{A}-V_{B}\) is
1 \(\mathrm{CV}\)
2 \(4 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(-2 \mathrm{~V}\)
Explanation:
Apply Ohm's law.
Current Electricity
268424
A uniform wire of resistance \(20 \Omega\) having resistance \(1 \Omega\) / m is bent in the form of circle as shown in fig. If the equivalent resistance between \(M\) and \(N\) is \(1.8 \Omega\), then the length of the shorter section is
1 \(2 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(1.8 \mathrm{~m}\)
4 \(18 \mathrm{~m}\)
Explanation:
Let the resistance of shorter part \(\mathrm{MN}\) bex. Then resistance of longer part is \((20-x) \Omega\) \(R_{e q}=\frac{(20-x) x}{20-x+x}=1.8 \Omega\) Solving we get \(x=2 \Omega\) So length of shorter part \(=2 \mathrm{~m}\)
Current Electricity
268425
If the voltmeter reads \(0.2 \mathrm{~V}\) and the ammeter reads \(0.101 \mathrm{~A}\), the resistance of the voltmeter is(in ohm)
1 500
2 1000
3 200
4 400
Explanation:
\(\left[\frac{2 x}{2+x}\right](0.101)=0.2 \quad\) solve for ' \(x\) '.
Current Electricity
268426
In the en then when both switches \(S_{1}, S_{2}\) are closed or opened. The value of resistance \(R\) is
1 \(200 \Omega\)
2 \(100 \Omega\)
3 \(400 \Omega\)
4 \(-300 \Omega\)
Explanation:
When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) areopened, \(\mathrm{i}=\frac{1.5}{450}\) When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) are closed, \(i=\frac{1.5[100+R]}{400 R+30,000}\)
Current Electricity
268427
In the following diagram ammeter reading is \(4 \mathrm{~A}\), voltmeter reading is \(20 \mathrm{~V}\), the value of \(R\) is
1 \(\lt 5 \Omega\)
2 \(<5 \Omega\)
3 \(=5 \Omega\)
4 \(\leq 5 \Omega\)
Explanation:
\(\mathrm{iR}=20\), here \(\mathrm{i}<4 \mathrm{~A} \Rightarrow \mathrm{R}\lt 5 \Omega\)
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Current Electricity
268423
In the circuit shown below, the cell has an emf of \(10 \mathrm{~V}\) and internal resistance of \(1 \mathrm{ohm}\) the other resistances are shown in the figure. The potential difference \(V_{A}-V_{B}\) is
1 \(\mathrm{CV}\)
2 \(4 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(-2 \mathrm{~V}\)
Explanation:
Apply Ohm's law.
Current Electricity
268424
A uniform wire of resistance \(20 \Omega\) having resistance \(1 \Omega\) / m is bent in the form of circle as shown in fig. If the equivalent resistance between \(M\) and \(N\) is \(1.8 \Omega\), then the length of the shorter section is
1 \(2 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(1.8 \mathrm{~m}\)
4 \(18 \mathrm{~m}\)
Explanation:
Let the resistance of shorter part \(\mathrm{MN}\) bex. Then resistance of longer part is \((20-x) \Omega\) \(R_{e q}=\frac{(20-x) x}{20-x+x}=1.8 \Omega\) Solving we get \(x=2 \Omega\) So length of shorter part \(=2 \mathrm{~m}\)
Current Electricity
268425
If the voltmeter reads \(0.2 \mathrm{~V}\) and the ammeter reads \(0.101 \mathrm{~A}\), the resistance of the voltmeter is(in ohm)
1 500
2 1000
3 200
4 400
Explanation:
\(\left[\frac{2 x}{2+x}\right](0.101)=0.2 \quad\) solve for ' \(x\) '.
Current Electricity
268426
In the en then when both switches \(S_{1}, S_{2}\) are closed or opened. The value of resistance \(R\) is
1 \(200 \Omega\)
2 \(100 \Omega\)
3 \(400 \Omega\)
4 \(-300 \Omega\)
Explanation:
When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) areopened, \(\mathrm{i}=\frac{1.5}{450}\) When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) are closed, \(i=\frac{1.5[100+R]}{400 R+30,000}\)
Current Electricity
268427
In the following diagram ammeter reading is \(4 \mathrm{~A}\), voltmeter reading is \(20 \mathrm{~V}\), the value of \(R\) is
1 \(\lt 5 \Omega\)
2 \(<5 \Omega\)
3 \(=5 \Omega\)
4 \(\leq 5 \Omega\)
Explanation:
\(\mathrm{iR}=20\), here \(\mathrm{i}<4 \mathrm{~A} \Rightarrow \mathrm{R}\lt 5 \Omega\)
268423
In the circuit shown below, the cell has an emf of \(10 \mathrm{~V}\) and internal resistance of \(1 \mathrm{ohm}\) the other resistances are shown in the figure. The potential difference \(V_{A}-V_{B}\) is
1 \(\mathrm{CV}\)
2 \(4 \mathrm{~V}\)
3 \(2 \mathrm{~V}\)
4 \(-2 \mathrm{~V}\)
Explanation:
Apply Ohm's law.
Current Electricity
268424
A uniform wire of resistance \(20 \Omega\) having resistance \(1 \Omega\) / m is bent in the form of circle as shown in fig. If the equivalent resistance between \(M\) and \(N\) is \(1.8 \Omega\), then the length of the shorter section is
1 \(2 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(1.8 \mathrm{~m}\)
4 \(18 \mathrm{~m}\)
Explanation:
Let the resistance of shorter part \(\mathrm{MN}\) bex. Then resistance of longer part is \((20-x) \Omega\) \(R_{e q}=\frac{(20-x) x}{20-x+x}=1.8 \Omega\) Solving we get \(x=2 \Omega\) So length of shorter part \(=2 \mathrm{~m}\)
Current Electricity
268425
If the voltmeter reads \(0.2 \mathrm{~V}\) and the ammeter reads \(0.101 \mathrm{~A}\), the resistance of the voltmeter is(in ohm)
1 500
2 1000
3 200
4 400
Explanation:
\(\left[\frac{2 x}{2+x}\right](0.101)=0.2 \quad\) solve for ' \(x\) '.
Current Electricity
268426
In the en then when both switches \(S_{1}, S_{2}\) are closed or opened. The value of resistance \(R\) is
1 \(200 \Omega\)
2 \(100 \Omega\)
3 \(400 \Omega\)
4 \(-300 \Omega\)
Explanation:
When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) areopened, \(\mathrm{i}=\frac{1.5}{450}\) When \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) are closed, \(i=\frac{1.5[100+R]}{400 R+30,000}\)
Current Electricity
268427
In the following diagram ammeter reading is \(4 \mathrm{~A}\), voltmeter reading is \(20 \mathrm{~V}\), the value of \(R\) is
1 \(\lt 5 \Omega\)
2 \(<5 \Omega\)
3 \(=5 \Omega\)
4 \(\leq 5 \Omega\)
Explanation:
\(\mathrm{iR}=20\), here \(\mathrm{i}<4 \mathrm{~A} \Rightarrow \mathrm{R}\lt 5 \Omega\)
