A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.
[Figure]
[Figure]
Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors.
[Figure]
Maximum current through resistor A is \( \sqrt{\dfrac{18}{2}} = 3 \text{ A} \).
Thus the maximum current through resistors B and C each is \( 3 \times \dfrac{1}{2} = 1.5 \text{ A} \).
Should the resistance of an ammeter be low or high? Give reason.
The resistance of an ammeter should be as low as possible. Ideally, it should be zero ohm so that it does not affect the current in the circuit.
Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4 Ω resistors? Give reason.
[Figure]
[Figure]
Yes. The total resistance of the parallel combination is 2 Ω, so the potential difference across the 2 Ω resistor is the same as that across the parallel combination.
How does use of a fuse wire protect electrical appliances?
If a current larger than a specified value flows through a circuit, the temperature of the fuse wire increases to its melting point. The fuse melts and the circuit breaks, protecting the appliances.
What is electrical resistivity? In a series electrical circuit comprising a resistor made up of metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?
Electrical resistivity \( \rho \) is the resistance of a wire of unit length and unit cross-section. When the length of the wire is doubled, resistance becomes \( R = \rho \dfrac{l}{A} \) and thus doubles. Since \( V = RI \) remains unchanged, current becomes half.