For Question 6-8
In the circuit shown on the right, the reading of the ammeter
is 1A when K is open.
When K is open, the circuit is equivalent to the one on the
right.
The current is 1A as read from the ammeter. Then the p.d.
across the resistor is 1A x 10 = 10V. The is also the e.m.f of
the Power supply. The e.m.f of the power supply will not change.
6.
What will the ammeter reading be when k is closed?.
If k is closed now, the p.d. across this 10 resistor is unchanged. Since I =V/R,
V is 10V and R is 10, current I through the ammeter is still 10V/10 = 1A.
Answer : 1A
7.
If the ammeter near the 10 resistor is removed and then connected between
X and Y, what will the ammeter reading be when K is closed.
Since K is closed , there will be a current passes through the 40 resistor.
Since I = V/R , I = 10V / 40 = 0.25 A. This 0.25A and the 1A currents will
combine and pass through the ammeter. Thus the answer is 1A+0.25A =
1.25A.
Answer = 1.25A
8.
What is the e.m.f of the power supply?
As calculated in question 6, the e.m.f. of the power supply is 10V.
Answer = 10V
For Questions 9-10
R
In the circuit on the right, when only switch P is closed, the
voltmeter reading is 4 V. But when only switch Q is closed, the
voltmeter reading is 8 V.
2
P
8
Q
V
When only P is closed, the circuit is equal to the one below.
Since the voltmeter reads 4V. The current
R
passes through the 2 resistor is I = V/R
= 4V/2= 2 A.
2
The p.d. across the resistor R is V = 2A x
R =2R (unit :V )
V
Then the e.m.f of the battery is
e.m.f. = 4+2R ( unit : V)
When only Q is close the circuit is equal to the one below.
Since the voltmeter reads 8V. The current
R
passes through the 8 resistor is I = V/R
= 8V/8= 1 A.
8
The p.d. across the resistor R is V = 1A x
R =1R
( unit : V)
V
Then the e.m.f of the battery is
e.m.f. = 8+1R ( unit : V)
Qn 10.
Since e.m.f is the same, that means:
4+2R = 8+1R
R = 4 ()
Qn 9.
The e.m.f. of the power supply is
e.m.f = 8+4 = 12 V.