Treal tHe DalIel 4 The bulbs in the circuit shown are identical. answering all t

Question

Treal tHe DalIel 4 The bulbs in the circuit shown are identical. answering all the questions. Evplain your reasoning. a. Rank buibs 1-6 in order of brightness. b. Rank the voltages across the bulbs. Explain your reasoning. Write an equation that relates the voltage across bulbs 3,5, and 6 to the battery voltage. c. d. Bulb 1 is removed from its socket. Does the brightness of bulb 2 increase, decrease, or remain the same? Explain. i. E placresedecresor rmals the ii. Does the brightness of ili. Does the brightness of bulb 3 increase, decrease, or remain the same? Explain.

Explanation / Answer

a) The brightness of a bulb is defined as the power of the bulb which is given by

P = I*V = I2 *R

hence, more the current flowing through the bulb, more bright is the bulb.

from the figure,

we see that the bulb 6 is the single bulb across the battery. thus it has the hghest current passing through and correspondingly brightest bulb in the circuit.

let the current through the bulb 6 is I, which is maximum.

now the current through bulb 4 and 5 are equal and is I/2

also the current through bulb 3 is 2I/3 and current through (1 &2) is I/3

so we see that bulb 3 is brighter than bulb 1or 2 .

hence the order of brightness of bulbs is : 6, (4 and 5), 3, (2 and 1)

b) we know from the ohms law that

V = I*R

thus from the given circuit,, we calculate the voltage through each bulb as

Voltage across bulb 6 = RI = V say
the voltage across bulb 3 = 2IR/3 = 0.66 V
Voltage across bulb 4 = voltage across bulb 5 = IR/2 = 0.50V
also,

Voltage across bulb 2 = voltage across bulb 1 = IR/3 = 0.33 V

c) let us assume that  E is the battery voltage
then, we have
E = V (1 +0.66 +0.5 + 0.33) = 2.49 V

or
V = E/ 2.49
thus
Voltage across bulb 6 = RI = E/ 2.49
Voltage across bulb 5 = 0.50V = 0.2 E
and

Voltage across bulb 3 = 2IR/3 = 0.27 V

i) since the bulb 1 is removed from the socket, then no current passes through bulb 2 as it is connected along the bulb 1. so, bulb 2 is off.

ii) here the total resistance increases, so the current decreases through bulb 6. decrease in current means decrease in the brightness. so for bulb 6, the brightness decreases .

iii) the current through bulb 3 increases and is the same as in bulb 6 and hence its brightness increases .

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