7. consider the circuits shown on the right. a. Are the bulbs C, D, and E connec

Question

7. consider the circuits shown on the right.
a. Are the bulbs C, D, and E connected in series, parallel or neither? explain
b. rank the bulbs in order of brightness. explain your ranking.
c. How will the brightness of bulbs A and B change if bulb C is unscrewed? Will the results be different if bulb D or E is unscrewed instead? Explain.

8. consider the circuit shown on the right. rank the brightness of the bulbs in the circuit. use the symbols =, <,>. explain.

9. in the two batteries below the batteries and bulbs are identical. compare the current in the circuit on the left to the current in the circuit on the right.

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Explanation / Answer

7 (a)

The bulbs C,D, and E are connected in parallel. Because the positive terminal of all three bulbs are conneted at one single point and the negative terminal of all three bulbs are connected at one single point.

So, all the 3 bulbs are conneted between the same 2 points. So all will have equal potential difference. Hence they are connected in parallel.

7 (b) the brightness of a bulb depends on the power of the bulb.

All bulbs are identical.

Lets take the potential difference of the battery to 100 V.

and the resistance of each bulb be 30 ohm

Now , equivalent resistance of the whole circuit = 30/2 + 30/3       = 15 +10 = 25 ohm  

( equivalent resistance of n identical resistors in parallen is given by R/n , So A and B are parallel and C,D,E are also parallel)

Current flowing in the circuit = 100/25 = 4 A.

Current passing throught A and B = 4/2 = 2 A

Power of A and B = I2R= 30*4 = 120 W.

Current flowing through C,D,E = 4/3 A

Power through each = 30*16/9 = 53.33 W

So,

According to ranking

A=B > C=D=E

A and B will have same brightness, greater than C,D,E all of which have same brightness.

7 (c) when bulb C is unscrewed

New Net resistance of the circuit = 30/2 +30/2 = 30 ohm

Net current = 100/30 = 3.33 A

Power of A and B I2R = 30* (3.33/2)2= 83.31 W.

So the power of both A and B is reduced. Hence the brightness is also reduced.

The results would be same if D or E is unscrewed, as the equivalent resistance would have been same.

8) Again lets take potential difference be 100 and

individual resistance of each bulb = 30 ohm

Now, equivalent resistance of the circuit =

resistance of upper branch = 30 +   60*30/90 = 50

Net resistance of the circuit = 30*50/80 = 18.75 ohm

Current flowing in the circuit = 100/18.75 = 5.333 A

Current flowing through A =     5.3333 *50/80 =3.3333 A

power bulb A = 100*3.3333 = 333.33 W

Current flowing through B = 5.333*30/80 = 2 A

power of B = I2R = 4*30 = 120 W

Current flowing through C = 2* 60/90 = 1.333 A

Power = I2R = 1.3332*30 =53.33 W

Current flowing through D and E = 2-1.333 = 0.667 A

Power of D= 0.6672*30 =13.34 W = power of E

So ,

the brightness of the bulb will be more for bulb having more power rating

So

A >B >C >D=E (Ans)

9) Let the potential difference of the battery be V

and resisitance of each bulb = R

Now current of circuit 1 = V/R

Current of circuit 2 = V/(R+R+R) = V/3R

So the current in the circuit 1 is thrice of current of circuit 2.

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