Question

PLEASE ANSWER BOTH:)

Question 7: Which of the following is the correct ranking of the brightness of the light bulbs in the circuit shown below? A A) C > A = B >D = E B) C>D= E > A=B C) A = B > C>D=E D) D = E > C> A=B E) A = B >D = EC B Question 8: In the circuit shown in question 7, the battery supplies 12 V and voltage across bulb D is 3 V. What is the voltage across bulb C? A) 1.5 V B) 3 V C 6 V D) 9V E) 12 V

Answer 1

Assuming all the bulbs are identical, with equal resistances.

A)

Req of the branch containing A and B : R/2

Req of the branch containing C, D and E:

$R_{eq} = \frac{R*2R}{3R} = \frac{2R}{3}$

Req of the complete circuit:

$R_{eq} = \frac{R}{2} + \frac{2R}{3} = \frac{7R}{6}$

So, total current through the circuit: I

$I = \frac{6V}{7R}$

Now, current through Bulb A = Current through Bulb B = I/2 = 3V/7R = 0.43 V/R

Also, current through Bulb C:

$I_C = \frac{6V}{7R} * \frac{2R}{3R} = \frac{12V}{21R} = 0.57\frac{V}{R}$

Also, current through Bulb D = Current through Bulb E :

$I_D = I_E = \frac{6V}{7R} * \frac{R}{3R} = \frac{6V}{21R} = 0.28\frac{V}{R}$

So, now from above:

I_C > I_A = I_B > I_D = I_E

So, the order of the brightness of the bulb:

A) C > A = B > D = E

B)

Voltage across Bulb E = Voltage across Bulb D = 3 V because both have same resistances and are connected in series.

Also, Voltage across Bulb C = Voltage across Bulb D + Voltage across Bulb E = 3 + 3 = 6 V

C) 6 V