Three resistors, with substances 250\(\Omega\), 500\(\Omega\) and 1k\(\Omega\) are connected in series. A 6V battery is connected to either end of the combination. Calculate the pot3ential difference between the end of the 250\(\Omega\) resistor.
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Correct Answer: Option B
Explanation:
Using voltage - divider method
v1 = \(\frac{(R_1)}{R_T}\)VT
= \(\frac{250}{1750}\) x \(\frac{6}{1}\)
= 0.86V
Using voltage - divider method
v1 = \(\frac{(R_1)}{R_T}\)VT
= \(\frac{250}{1750}\) x \(\frac{6}{1}\)
= 0.86V