In the given circuit diagram the resistance R 1 = 8 Ω , R 2 = 4 Ω . The circuit is powered by a source of emf E = 48 V : If the contact is moved from 1 to 2 and key K is closed, the total current in the circuit changes by 6A. Calculate R 3 .

When the contact 2 is used and Key K is closed , R 1 = 8 Ω , R 2 = 4 Ω , R 3 is included in the circuit, R 1 , R 2 are in series and their combination is in parallel with R 3 Equivalent resistance is given by R = ( R 1 + R 2 ) R 3 R 1 + R 2 + R 3 = 12 R 3 12 + R 3 So,total current in the circuit will be I = E R = 48 12 + R 3 12 R 3 = 4 12 + R 3 R 3 = 12 A R 3 = 6 Ω

The circuit diagram shows a transistor as an amplifier : What is the phase difference between the input and the output voltage ?

In the given circuit diagram, Input part of the circuit is made by Base- Emitter of the transistor and Output part of the circuit is formed by Collector and Emitter . Since Emitter junction is common to both input and output parts hence this configuration is called CE amplifier circuit.The phase difference between the input and output voltage of CE amplifier circuit is 180 ∘

The circuit diagram shows a transistor as an amplifier : How does the collector current change when the base current increases ?

In the given circuit diagram, Input part of the circuit is made by Base- Emitter of the transistor and Output part of the circuit is formed by Collector and Emitter . Since Emitter junction is common to both input and output parts hence this configuration is called CE amplifier circuit.The phase difference between the input and output voltage of CE amplifier circuit is 180 ∘ An increase in base current means more bias to the transistor which further means collector current will also increase but till the limiting value of the transistor parameters.

In the given circuit diagram the resistance R 1 = 8 Ω , R 2 = 4 Ω . The circuit is powered by a source of emf E = 48 V : If the contact is moved from 1 to 2 and key K is closed, the total current in the circuit changes by 6A. Calculate R 3 .

When the contact 2 is used and Key K is closed , R 1 = 8 Ω , R 2 = 4 Ω , R 3 is included in the circuit, R 1 , R 2 are in series and their combination is in parallel with R 3 Equivalent resistance is given by R = ( R 1 + R 2 ) R 3 R 1 + R 2 + R 3 = 12 R 3 12 + R 3 So,total current in the circuit will be I = E R = 48 12 + R 3 12 R 3 = 4 12 + R 3 R 3 = 12 A R 3 = 6 Ω

The circuit diagram shows a transistor as an amplifier : How does the collector current change when the base current increases ?

In the given circuit diagram, Input part of the circuit is made by Base- Emitter of the transistor and Output part of the circuit is formed by Collector and Emitter . Since Emitter junction is common to both input and output parts hence this configuration is called CE amplifier circuit.The phase difference between the input and output voltage of CE amplifier circuit is 180 ∘ An increase in base current means more bias to the transistor which further means collector current will also increase but till the limiting value of the transistor parameters.

The circuit diagram shows a transistor as an amplifier : What is the phase difference between the input and the output voltage ?

In the given circuit diagram, Input part of the circuit is made by Base- Emitter of the transistor and Output part of the circuit is formed by Collector and Emitter . Since Emitter junction is common to both input and output parts hence this configuration is called CE amplifier circuit.The phase difference between the input and output voltage of CE amplifier circuit is 180 ∘

The circuit diagram shows a potentiometer for determining the internal resistance of cell ε 1 With the key ‘K’ open, the balancing length obtained is AD : If the key ‘K’ is closed, towards which end of the wire will the balance point ‘D’ shift ? Why ?

For the given experimental setup, AD is obtained as the balancing length as the potential difference across AD is equivalent to the emf of the cell to be checked. When the Key is closed , Resistance R is included in the circuit and hence , the terminal potential difference across cell increases and hence balancing length increases. So, the balance point is shifted to the right.

The circuit diagram shows a potentiometer for determining the internal resistance of cell ε 1 With the key ‘K’ open, the balancing length obtained is AD : If the resistance R 1 is increased, towards which end of the wire will the balance point ‘D’ shift ? Why ?

For the given experimental setup, AD is obtained as the balancing length as the potential difference across AD is equivalent to the emf of the cell to be checked. As we increase R 1 current flowing in the arm AB is decreased, hence the same potential will be dropped on a higher resistance value, hence D point will shift towards B end.