A battery with emf 10.30 V and internal resistance 0.50 Ω is inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-V battery. What is the potential difference \( V_{bc} \) between the terminals of the 4.00-V battery now?

The potential difference across the 4.00 V battery, Vbc, in a circuit where it is in parallel with a series combination of 10.30 V and 8.00 V batteries, is equal to the total emf of the two batteries in series minus the potential drop due to their current through the 0.50 Ω internal resistance.

Explanation:

To determine the potential difference Vbc across the terminals of the 4.00 V battery, we first need to understand the circuit's behavior. The circuit now consists of two batteries connected in series - one with 10.30 V (internal resistance 0.50 Ω) and another of 8.00 V - and the 4.00 V battery is in parallel with this series combination.

Because the batteries in series share the same current, we can calculate the total current (I) using Ohm's law and considering the total emf and resistance:

I= (10.30 V + 8.00 V)/( R + 0.50 Ω)

Note that R includes any circuit resistance (which hasn't been mentioned) and the 4.00 V battery's own internal resistance. Therefore, Vbc, the potential difference across the 4.00 V battery will be equal to the total potential provided by the series batteries minus the potential drop due to their current through the 0.50 Ω internal resistance.

Vbc = (10.30 V + 8.00 V) - I * 0.50 Ω

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