As per the given question, the following can be determined:

Effective resistance of two 8 Ω resistors in combination
⇒ R = \(\frac{R_1 R_2}{R_1 R_2} = \frac{8\ \times \ 8}{8\ + \ 8}\) = 4 Ω
Current that flows through 4Ω resistor
⇒ \(I = \frac{V}{R}= \frac{8}{4 + \frac{8 \times 8}{8+8}}\)
Potential difference across 4Ω resistance
⇒ V = IR = 1 × 4 = 4V
Disspated Power in 4 Ω resistor
⇒ P = I2R = 12 × 4 = 4W
Difference, if any, in ammeter readings
There is zero difference as because no current is seen to flow through the elements in a series circuit.
Check More: Class 12 Physics Notes
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