Question 569: Current Electricity

Question

A circular coil of radius $$R$$ has a resistance of $$40 \Omega$$. Figure shows two points $$P$$ and $$Q$$ on the circumference separated by a distance $$\frac{\pi R}{2}$$ which are connected to a 16 V battery with internal resistance of $$0.5 \Omega$$. What is the value of current $$I$$ flowing through the circuit?

Accepted Answer

First, calculate the total resistance between points $ P $ and $ Q $ :
$$ \begin{aligned} R_{PQ} & = \left( \frac{30 \Omega 	imes 10 \Omega}{30 \Omega + 10 \Omega} ight) + 0.5 \Omega \\        & = \frac{300 \Omega}{40 \Omega} + 0.5 \Omega \\        & = 7.5 \Omega + 0.5 \Omega \\        & = 8 \Omega \end{aligned} $$
Next, determine the current flowing through the circuit using Ohm's law : $ V = IR $
$$ I = \frac{V}{R_{PQ}} $$
Where $ V $ is the voltage of the source. Therefore,
$$ I = \frac{16 	ext{ V}}{8 \Omega} = 2 	ext{ A} $$
The current $ I $ flowing through the circuit is thus 2 A.

Answer

3A

Answer

2A

Answer

1A

Answer

0.5A

Current Electricity

A circular coil of radius $R has a resistance of 40 \Omega. Figure shows two points P and Q on the circumference separated by a distance \frac{\pi R}{2} which are connected to a 16 V battery with internal resistance of 0.5 \Omega. What is the value of current I$ flowing through the circuit?

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