Current Electricity Test

Result

Current Electricity Test - 80

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Weekly Quiz Competition

Question 1
1 / -0

Calculate the potential difference between, points A and B current flowing in $$10\Omega $$ resistor in the part of the network below:-

A
$$20 V,2 A$$

B
$$50 V, 1 A$$

C
$$40 V, 1 A$$

D
$$30 V,1 A$$

Solution
Question 2
1 / -0

A part of circuit with current is shown. The value of $$I $$ is :

A
$$1A$$

B
$$2A$$

C
$$4A$$

D
Zero

Solution

According to KCL the total current that enters ie equal to the total current living from the node. Hence the current in the shown part is zero.
Question 3
1 / -0

The potential difference across the terminals of a battery is $$10V$$ when there is a current of $$3A$$ in the battery from the negative to the positive terminal. When the current is $$2A$$ in the reverse direction, the potential difference becomes $$15V$$. The internal resistance of the battery is?

A
$$ 2.5\Omega$$

B
$$3.2\Omega$$

C
$$1.5\Omega$$

D
$$4\Omega$$

Solution
Question 4
1 / -0

In the circuit shown, the current in the $$1\ \Omega$$ resistor is :

A
$$0.13\ A$$, from $$Q$$ to $$P$$

B
$$0.75\ A$$, from $$P$$ to $$Q$$

C
$$1.3\ A$$, from $$P$$ to $$Q$$

D
$$0\ A$$

Solution
Question 5
1 / -0

The equivalent resistance between points $$A$$ and $$B$$ is

A
$$\dfrac {91}{2}\Omega$$

B
$$\dfrac {65}{2}\Omega$$

C
$$\dfrac {45}{2}\Omega$$

D
$$\dfrac {5}{2}\Omega$$

Solution
Question 6
1 / -0

In the circuit shown below, the cell is ideal, with $$emf=2V$$. The resistance of the coil of the galvanometer $$G$$ is $$1 \Omega$$. The steady current through the galvanometer is

A
$$0.2A$$

B
$$0.1A$$

C
$$0.4A$$

D
Zero

Solution
Question 7
1 / -0

If the current in the toroidal solenoid increases uniformly from zero to $$6.0 A$$ in $$3.0\mu s$$.Self inductance of the toroidal solenoid is $$40 \mu H$$. The magnitude of self-induced emf is

A
$$24 V$$

B
$$48 V$$

C
$$80 V$$

D
$$160 V$$

Solution
Question 8
1 / -0

In the determination of the internal resistance of a cell using a potentiometer, when the cell is shunted by a resistance $$"R"$$ and connected in the secondary circuit, the balance length is found to be $$L_1$$. On doubling the shunt resistance, the balance length is found to increase to $$L_2$$. The value of the internal resistance is:

A
$$\dfrac{2 R(L_2 - L_1)}{(L_1 - 2 L_2)}$$

B
$$\dfrac{2 R(L_2 - L_1)}{(2 L_1 - L_2)}$$

C
$$\dfrac {R(L_2 - L_1)}{(L_1 - 2 L_2)}$$

D
$$\dfrac{R(L_2 - L_1)}{(2 L_1 - L_2)}$$

Solution
Question 9
1 / -0

In the adjoining figure the equivalent resistance between $$A$$ and $$B$$ is

A
$$5\ \Omega $$

B
$$8\ \Omega $$

C
$$2.5\ \Omega $$

D
$$6.8\ \Omega $$

Solution

We are given circuit, we have to calculate equivalent between A and B.
Clearly in 2nd fig.
2, 2 and 2 are in series
So, $$R_1 = 2 + 2 + 2$$
$$= 6\Omega$$
Now, $$6\Omega$$ and $$6\Omega$$ are in parallel
$$\therefore R_2 = \dfrac{6\times 6}{12} = 3\Omega$$
Now, Equivalent Resistance ($$R_{AB}$$) = 1 + 3 + 1
$$= 5\Omega$$
Hence option A is correct.
Question 10
1 / -0

A horizontal straight conductor of length $$l$$ placed along east-west direction falls under gravity from height $$h$$ at a place where horizontal component of magnetic field is $$H$$ and vertical component of magnetic field is $$V$$. Then :

A
No electromotive force is induced along the length of the conductor

B
An electromotive force is induced along the length which has maximum value $$Hl\sqrt{2gh}$$

C
An electromotive force is induced along the length which has maximum value $$Vl\sqrt{2gh}$$

D
An electromotive force is induced along the length which has maximum value $$l\sqrt{2gh (V^2 + H^2)}$$

Solution