Example 1. If bulb rating is 100 watt and 220 V then determine
(a) Resistance of filament
(b) Current through filament
(c) If bulb operate at 110 volt power supply then find power consume by bulb.
Sol. Bulb rating in 100 W and 220 V bulb means when 220 V potential difference is applied between the two ends then the power consume is 100 W
Here V = 220
P = 100
So R = 484 W
Since Resistance depends only on material hence it is constant for bulb
power consumed at 110 V
Therefore, power consumed =
Example 3. In the following figure, grade the bulb in order of their brightness :
Sol.
Power = i^{2}R
As current passing through every bulb is same
Therefore, Brightness order is B_{3} > B_{2} > B_{1}
Example 4.
The above configuration shows three identical bulbs, Grade them in order of their brightness.
Sol. B_{1} & B_{2} withdraw less current as compared to B_{3} because in series they give 2R resistance where as R is the resistance due to B_{3}.
Power = i^{2}R
Therefore, Brightness order : B_{3} > B_{2} = B_{1}.
Example 5.
Grade the bulbs in order of their brightness (All bubls are identical)
Sol.
Therefore, Order of Brightness : B_{5} > B_{1} = B_{2} > B_{6} > B_{4} = B_{3}
(B) Maximum power transfer theorem
Let E be emf and r internal resistance of the battery. It is supplying current to an external resistance R
current in circuit I =
The power absorbed by load resistor R is
P = I^{2}R =
For maximum power transfer we take the derivative of P w.r.t R, set it equal to zero and solve the equation for R.
Solving for R, we have
(R r)^{2}  R (2) (R r) = 0
(R r)  2R = 0
R = r
For a given real battery the load resistance maximizes the power if it is equal to the internal resistance of the battery.
The maximum power transfer theorem in general, holds for any real voltage source. The resistance R may be a single resistor or R may be the equivalent resistance of a collection of resistors.
Regarding an ammeter it is worth noting that :
Example 6. What is the value of shunt which passes 10% of the main current through a galvanometer of 99 ohm?
Sol. As in figure R_{g}I_{g} = (I  I_{g})S
⇒
⇒ S = 11 ?
For calculation it is simply a resistance
Resistance of ammeter
for S << R_{G} ⇒ R_{A} = S
Example 7. Find the current in the circuit also determine percentage error in measuring in current through an ammeter (a) and (b).
Sol. In A
In B
Percentage error is = = 20% Ans.
Here we see that due to ammeter the current has reduced. A good ammeter has very low resistance as compared with other resistors, so that due to its presence in the circuit the current is not affected.
Example 8. Find the reading of ammeter. Is this the current through 6 W ?
Sol.
Current through battery
So, current through ammeter
No, it is not the current through the 6W resistor.
Note: Ideal ammeter is equivalent to zero resistance wire for calculation potential difference across it is zero.
It is a device used to measure potential difference and is always put in parallel with the 'circuit element' across which potential difference is to be measured e.g., in Figure (A) voltmeter V_{1} will measure potential difference across resistance R_{1}, V_{2} across resistance R_{2} and V across (R_{1} R_{2}) with V = V_{1} V_{2}
Regarding a voltmeter it is worth noting that :
(1)The reading of a voltmeter is always lesser than true value, e.g., if a current I is passing through a resistance R [Fig. (B)], the true value V = IR. However, when a voltmeter having resistance r is connected across R, the current through R will become
and so
and as voltmeter is connected across R its reading V' is lesser than V.
(2) Greater the resistance of voltmeter, more accurate will be its reading. A voltmeter is said to be ideal if its resistance r is infinite, i.e., it draws no current from the circuit element for its operation. Ideal voltmeter has been realised in practice in the form of potentiometer.
(3) To convert a galvanometer into a voltmeter of certain range say V, a high resistance R is connected in series with the galvanometer so that current passing through the galvanometer of resistance G becomes equal to its full scale deflection value I_{g}. This is possible only if
V = I_{g} (G R) i.e.,
Example 9. A voltmeter has a resistance of G ohm and range of V volt. Calculate the resistance to be used in series with it to extend its range to nV volt.
Sol. Full scale current i_{g} =
to change its range
V_{1} = (G R_{s}) i_{g}
⇒ nV = (G R_{s})
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