Find out the following from the circuit given (a) Potential difference across 4 ohm resistor. (b) Power dissipated in 4 ohm resistor. (c) Difference in ammeter readings, if any, of A1 and A2.
Dear Student
Two 8 ohms resistance are connected in parallel so they can be replaced by equivalent one 4 ohm resistance which is connected to another 4 ohm resistance in series.
So total resistance in circuit is 4 ohm + 4 ohm = 8 ohm
Now the Ohm's law equation (V = I . R) can be used to determine the total current in the circuit. In doing so, the total resistance and the total voltage (or battery voltage) will have to be used.
The 1 Amp current represents the current at the battery location. The current in series-connected resistors is everywhere the same. Thus,
Current across 4 ohm resistor is equal to 1 A.
For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I8 + I8 must equal 1 Amp.
Since the resistance values are equal, the current values in these two resistors are also equal. Therefore, the current in resistors 8 ohm and 8 ohm connected in parallel are both equal to 0.5 Amp.
Now voltage drop across 4 ohm resistance = I4 .R = (1 Amp)(4 ohm) = 4 Volt.
Power dissipated across 4 ohm resistor P= IV = (1 Amp)(4 V) = 4 Watt.
Both ammeter will read same readings because the sum of the current in each individual branch is equal to the current outside the branches.
Regards
Two 8 ohms resistance are connected in parallel so they can be replaced by equivalent one 4 ohm resistance which is connected to another 4 ohm resistance in series.
So total resistance in circuit is 4 ohm + 4 ohm = 8 ohm
Now the Ohm's law equation (V = I . R) can be used to determine the total current in the circuit. In doing so, the total resistance and the total voltage (or battery voltage) will have to be used.
The 1 Amp current represents the current at the battery location. The current in series-connected resistors is everywhere the same. Thus,
Current across 4 ohm resistor is equal to 1 A.
For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I8 + I8 must equal 1 Amp.
Since the resistance values are equal, the current values in these two resistors are also equal. Therefore, the current in resistors 8 ohm and 8 ohm connected in parallel are both equal to 0.5 Amp.
Now voltage drop across 4 ohm resistance = I4 .R = (1 Amp)(4 ohm) = 4 Volt.
Power dissipated across 4 ohm resistor P= IV = (1 Amp)(4 V) = 4 Watt.
Both ammeter will read same readings because the sum of the current in each individual branch is equal to the current outside the branches.
Regards