Question

The bulbs in the given circuit are identical and the battery is ideal. Part a onsider the circuit as shown. Figure 2 i. Rank bulbs A, B, and C in order of brightness. Explain how you determined your answer. ii. Rank the voltages across the bulbs. Explain. iii. Write an equation that relates the voltage across bulbs A and B to the battery voltage. iv. Is the voltage across bulb A greater than, less than, or equal to one half the battery voltage? Explain your easoning

Answer 1

The brightness of the bulbs is directly proportional to the power consumed by them. The bulb consuming the most power is the brightest. The power consumed by a bulb= i² * R where i is the current through the bulb and R is its resistance.

Part a)

i)

A>B=C (The order of the brightness

The current in the bulb A is the greatest so it consumes the greatest power. Hence it is the brightest.

The current through B and C is the same, so they have the same brightness.

ii)

A>B =C (the order of the voltage across the bulbs)

The bulbs are identical, so the voltage across them is dictated by the current through them, Sice the current through A is the greatest, Hence the voltage across A is the greatest. The currents in B and C is the same. hence they have the same voltage across them.

iii) The sum of the voltages across a closed path = 0 (Kirchhoff's law)

so V + (-V_A) + (-V_B ) = 0 (negative sign is due voltage drops across A and B

so V = V_A + V_B (V is the battery voltage)

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iv) Before adding C in parallel to B, the voltage across A is half the battery voltage.

Now after adding C the resistance of the circuit decreases, so the current through the circuit increases. So the voltage across A is now greater than 1/2 the battery voltage.