Solved current electricity numerical for class 10

CBSE ADDA
1. Calculate the resistance of a copper wire of length 1 m and area of cross section 2 mm2. Resistivity of copper is1710−8mn

Solution : R=1A=(1710−8m)1m2(10−3m)=8510−3
2.Calculate the potential difference across each resistor in the circuit shown in Figure 5.W2.
Solution : The three resistors are joined in series. Their equivalent resistance is
Req=4+6f +10=20
The current through the cell is i = V520=025A
The same current passes through each resistor. 

Using Ohm's law,

the potential difference across the 4-£1 resistor = 025A4=1V
across the 6-Q resistor = 025A6=15V and across the 10-0 resistor = 025A10=25V
3.

Consider the circuit shown in Figure 5.W4. The voltmeter on the left reads 10 V and that on the right reads 8 V.

 Find (a) the current through the resistance R, 

(b) the value of K and 

(c) the potential difference across the battery.
Solution : (a) Apply Ohm's law to the 4-Q resistor. 

The current through this resistor is i=V84=2A
As the two resistors are connected in series, the same current passes through the two resistors (the voltmeters draw only a negligible current). Hence, the current through R is 2 A,
(b) Applying Ohm's law to the resistance R,
10 V = R x (2 A) or R=01V2A=5
(c) The potential difference across the battery is
VA−VC=(VA−VB)+(VB−VC)=10V+8V=18V
4. When two resistors are joined in series, the equivalent resistance is 90 ft. When the same resistors are joined in parallel, the equivalent resistance is 20 ft. Calculate the resistances of the two resistors.
Solution : R1+R2=90 ... (i) and R1R2R1+R2=20 ....(ii)
R1R2=(R1+R2)(20)=(90)(20)=18002
Using the relation (a−b)2=(a+b)2−4ab
    1−R2=(R1+R2)2−4R1R2
=81002−72002  =9002=30 ....(iii)
From (i) and (iii), R1=60R2=30
Q.A current of 4 A passes through a resistance of 100 ft for 15 minutes. Calculate the heat produced in calories.

Solution : The heat produced is u=i2Rt =(4A)2(100)(1560s)=144106J
Now 4.186 J = 1 cal.
Thus 144106J=4186144106cal=34105cal