Current Electricity Test

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Current Electricity Test - 78

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Question 1
1 / -0

what is $$i$$?

A
$$-4$$mA

B
$$+2$$mA

C
$$+4$$mA

D
$$+8$$mA

Solution
Question 2
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A potentiometer wire of length $$L$$ and a resistance $$r$$ are connected in series with a battery of e.m.f. $$E_0$$ and a resistance $$r_1$$. An unknown e.m.f. $$E$$ is balanced at a length $$l$$ of the potentiometer wire. The e.m.f. $$E$$ will be given by:

A
$$\displaystyle \frac{LE_0r}{(r+r_1)l}$$

B
$$\displaystyle \frac{LE_0 r}{lr_1}$$

C
$$\displaystyle \frac{E_0 r}{(r+r_1)}. \frac{l}{L}$$

D
$$\displaystyle \frac{E_0l}{L}$$

Solution

Current in the circuit having $$E_0$$ is: $$i = \frac{E_0}{r + r_1}$$
Voltage dropped in potentiometer wire of length L is: $$ir = \frac{E_0 r}{r + r_1}$$
Since the balanced point is obtained at length l of the potentiometer wire, the emf of the cell E has to match the voltage dropped in length l of the wire.
$$\therefore E = \frac{E_0 r}{r + r_1} \times ( \frac{l}{L})$$
Question 3
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Arrange the order of power dissipated in the given circuits, if the same current enters at point $$A$$ in all the circuits and resistance of each resistor is $$R$$.

A
$$(ii) > (iii) > (iv) > (i)$$

B
$$(iii) > (ii) > (iv) > (i)$$

C
$$(iv) > (iii) > (ii) > (i)$$

D
$$(i) > (ii) > (iii) > (iv)$$

Solution
Question 4
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A cell of constant $$emf$$ first connected of a resistance $$R_{1}$$ and then connected to a resistance $$R_{2}$$. If power delivered in both cases is same then the internal resistance of the cell is:

A
$$\sqrt {R_{1}R_{2}}$$

B
$$\sqrt {\dfrac {R_{1}}{R_{2}}}$$

C
$$\dfrac {R_{1} - R_{2}}{2}$$

D
$$\dfrac {R_{1} + R_{2}}{2}$$

Solution

Current given by cell
$$I = \dfrac {E}{R + r}$$
Power delivered in first case
$$P_{1} = I^{2}R_{1}$$
$$= \left (\dfrac {E}{R_{1} + r}\right )^{2} R_{1}$$
Power delivered in second case
$$P_{2} = I^{2}R_{2}$$
$$= \left (\dfrac {E}{R_{2} + r}\right )^{2}R_{2}$$
Power delivered is same in the both the cases
$$\left (\dfrac {E}{R_{1} + r}\right )^{2} R_{1} = \left (\dfrac {E}{R_{2} + r}\right )^{2} R_{2}$$
$$\dfrac {R_{1}}{(R_{1} + r)^{2}} = \dfrac {R_{2}}{(R_{2} + r)^{2}}$$
$$R_{1}(R_{2}^{2} + r^{2} + 2R_{2}r) = R_{2}(R_{1}^{2} + r^{3} + 2R_{1}r)$$
$$R_{1}R_{2}^{2} + R_{1}r^{2} + 2R_{1}R_{2}r = R_{2}R_{1}^{2} + R_{2}r^{2} + 2R_{1}R_{2}r$$
$$R_{1}R_{2}^{2} - R_{2}R_{1}^{2} = R_{2}r^{2} - R_{1}r^{2}$$
$$R_{1}R_{2}(R_{2} - R_{1}) = r^{2} (R_{2} - R_{1})$$
$$r = \sqrt {R_{1} R_{2}}$$
Question 5
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Potentiometer wire of length $$1$$m is connected in series with $$490\Omega$$ resistance and $$2$$V battery. If $$0.2$$mV/cm is the potential gradient, then resistance of the potentiometer wire is?

A
$$4.9\Omega$$

B
$$7.9\Omega$$

C
$$5.9\Omega$$

D
$$6.9\Omega$$

Solution

$$R_p$$ is resistance of potentiometer.

Potential difference across potentiometer=potential gradient $$\times $$ length

$$=0.2\times 10^{-3}\times 100=0.02V$$

Current in circuit$$=I=\dfrac{E}{R+R_p}=\dfrac{2}{490+R_p}$$

Potential difference across potentiometer=$$IR_p$$

$$\implies \dfrac{2R_p}{490+R_p}=0.02$$

$$\implies 1.98R_p=9.8$$

$$\implies R_p=4.9\Omega$$

Answer-(A)
Question 6
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Two wire of same metal have same length but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thick wire is $$10\Omega$$. The total resistance of combination will be

A
$$10\Omega$$

B
$$20\Omega$$

C
$$40\Omega$$

D
$$100\Omega$$

Solution
Question 7
1 / -0

Three equal resistors connected in series across a source of emf together dissipate $$10$$ watt. If the same resistors are connected in parallel across the same emf, then the power dissipated will be:

A
$$10\ watt$$

B
$$30\ watt$$

C
$$\dfrac{10}{3}\ watt$$

D
$$90\ watt$$

Solution
Question 8
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A cell of emf $$E$$ is connected to a resistance $${R}_{1}$$ for time $$t$$ and the amount of heat generated in it is $$H$$. If the resistance $${R}_{1}$$ is replaced by another resistance $${R}_{2}$$ and is connected to the cell for the same time $$t$$, the amount of heat generated in $${R}_{2}$$ is $$4H$$. Then internal resistance of the cell is:

A
$$\cfrac { 2{ R }_{ 1 }+{ R }_{ 2 } }{ 2 } $$

B
$$\sqrt { { R }_{ 1 }{ R }_{ 2 } } \times \cfrac { 2\sqrt { { R }_{ 2 } } -\sqrt { { R }_{ 1 } } }{ \sqrt { { R }_{ 2 } } -2\sqrt { { R }_{ 1 } } } $$

C
$$\sqrt { { R }_{ 1 }{ R }_{ 2 } } \times \cfrac { \sqrt { { R }_{ 2 } } -2\sqrt { { R }_{ 1 } } }{ 2\sqrt { { R }_{ 2 } } -\sqrt { { R }_{ 1 } } } $$

D
$$\sqrt { { R }_{ 1 }{ R }_{ 2 } } \times \cfrac { \sqrt { { R }_{ 2 } } -\sqrt { { R }_{ 1 } } }{ \sqrt { { R }_{ 2 } } +\sqrt { { R }_{ 1 } } } $$

Solution

Heat generated is given as, $$H={ I }^{ 2 }R \times t $$
According to given condition:
$$\Rightarrow { I }_{ 1 }^{ 2 }{ R }_{ 1 } \times t =H$$ and $$ { I }_{ 2 }^{ 2 }{ R }_{ 2 } \times t=4H$$
$$\Rightarrow \cfrac { { E }^{ 2 } }{ { \left( { R }_{ 1 }+r \right) }^{ 2 } } { R }_{ 1 } \times t =H$$ and $$ \cfrac { { E }^{ 2 } }{ { \left( { R }_{ 2 }+r \right) }^{ 2 } } { { R }_{ 2 } } \times t =4H$$

$$\Rightarrow \cfrac { { R }_{ 2 } }{ { \left( { R }_{ 2 }+r \right) }^{ 2 } } =4\cfrac { { R }_{ 1 } }{ { \left( { R }_{ 1 }+r \right) }^{ 2 } } $$

$$\Rightarrow \sqrt { \left( { R }_{ 2 }+r \right) } \left( { R }_{ 1 }+r \right) =2\sqrt { { R }_{ 1 } } \left( { R }_{ 2 }+r \right) $$

On solving, $$r=\sqrt { { R }_{ 1 }{ R }_{ 2 } }\times \cfrac { \sqrt { { R }_{ 2 } } -2\sqrt { { R }_{ 1 } } }{ 2\sqrt { { R }_{ 2 } } -\sqrt { { R }_{ 1 } } } $$
Question 9
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A potentiometer (variable resistor) is connected in a simple circuit as shown below.
The student varies the resistance on the potentiometer and measures the corresponding current in the circuit. The table of data and the resistance vs. current graph created by the student is shown below.
$$R$$ (Ohm)

$$I$$ (A)
0.5 5.99
1.0 3.02
1.5 2.07
2.0 1.56
2.5 1.19
3.0 1.04
3.5 0.83
4.0 0.72
4.5 0.70
5.0 0.62
5.5 0.53
6.0 0.49
6.5 0.45
8.5 0.36
9.0 0.33
9.5 0.31
10.00 0.29
Based on the experimental data collected by the student, determine the emf of the battery.

A
0.33 V

B
0.5 V

C
1.0 V

D
2.0 V

E
3.0 V

Solution

$$E=IR$$

For all the data, we see that $$E\approx 3V$$

Answer-(E)
Question 10
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The temperature coefficient of resistance of a wire is $$\displaystyle { 0.00125 }/{ ^{ \circ }{ C } }$$. Its resistance is $$\displaystyle 1\Omega $$ at $$300\ K$$. Its resistance will be $$\displaystyle 2\ \Omega$$ at:

A
$$1127\ K$$

B
$$1400\ K$$

C
$$1154\ K$$

D
$$1100\ K$$

Solution

Given : $$\alpha = 0.00125/^oC$$
Using $$R_T = R_{273} (1 + \alpha (T- 273) )$$
For $$T = 300$$, $$R_{300} = 1\Omega$$
$$\therefore$$   $$R_{300} = R_{273} (1 + \alpha (300- 273) )$$
OR   $$1 = R_{273} (1 + 0.00125 (300- 273) )$$ $$\implies R_{273} = 0.967\Omega$$

Let at temperature be $$T$$ when $$R_T = 2\Omega$$
$$\therefore$$     $$R_T = R_{273} (1 + \alpha (T- 273) )$$
OR    $$2 = 0.967(1 + 0.00125 (T- 273) )$$
OR $$T- 273 = 854.6$$
$$\implies$$ $$T = 273 + 854.6 = 1127.6 K$$