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.
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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.
$I_{total }=\frac{V}{R_{total}}=\frac{8v}{8 ohm}= 1 Ampere$
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