(a) Given n resistors each of resistance R, how will you combinethem to get the (i) maximum (ii) minimum effective resistance?What is the ratio of the maximum to minimum resistance? (b) Given the resistances of 1 W, 2 W, 3 W, how will be combine themto get an equivalent resistance of (i) (11/3) W (ii) (11/5) W, (iii) 6W, (iv) (6/11) W?(c) Determine the equivalent resistance of networks shown inFig. 3.31.
a)
Given: The total number of resistors is n and the resistance of each resistor is R .
(i)
When all the resistors are connected in series, then the effective resistance is the maximum.
The expression for the maximum effective resistance is given as,
R 1 = n R (1)
Where, the maximum effective resistance is R 1 , the total number of resistors is n and the resistance of each resistor is R .
(ii)
When all the resistors are connected in parallel, then the effective resistance is the minimum.
The expression for the minimum effective resistance is given as,
R 2 = R n (2)
Where, the minimum effective resistance is R 2 , the total number of resistors is n and the resistance of each resistor is R .
By dividing the equation (1) by (2), we get
R 1 R 2 = n R ( R n ) = n 2 R R = n 2
Thus, the ratio of the maximum to the minimum resistance is n 2 .
b)
Given: The resistances are 1 Ω , 2 Ω and 3 Ω .
(i)
Consider the circuit combination for the required equivalent resistance is given as,
The equivalent resistance of the above circuit is given as,
R = ( R 1 R 2 R 1 + R 2 ) + R 3
Where, the equivalent resistance is R , the resistances connected in parallel are R 1 and R 2 , the resistance connected in series with R 1 and R 2 is R 3 .
By substituting the given values in the above expression, we get
R = 2 × 1 2 + 1 + 3 = 2 3 + 3 = 11 3 Ω
(ii)
Consider the circuit combination for the required equivalent resistance is given as,
The expression for the equivalent resistance of the above circuit is given as,
R = ( R 1 R 2 R 1 + R 2 ) + R 3
Where, the equivalent resistance is R , the resistances connected in parallel are R 1 and R 2 , the resistance connected in series with R 1 and R 2 is R 3 .
By substituting the given values in the above expression, we get
R = 2 × 3 2 + 3 + 1 = 6 5 + 1 = 11 5 Ω
(iii)
Consider the circuit combination for the required equivalent resistance is given as,
The expression for the equivalent resistance of the above circuit is given as,
R = R 1 + R 2 + R 3
Where, the equivalent resistance is R and the resistances connected in series are R 1 , R 2 and R 3 .
By substituting the given values in the above expression, we get
R = 1 + 2 + 3 = 6 Ω
(iv)
Consider the circuit combination for the required equivalent resistance is given as,
The expression for the equivalent resistance of the above circuit is given as,
1 R = 1 R 1 + 1 R 2 + 1 R 3
Where, the equivalent resistance is R and the resistances connected in parallel are R 1 , R 2 and R 3 .
By substituting the given values in the above expression, we get
1 R = 1 1 + 1 2 + 1 3 = 2 + 1 2 + 1 3 = 9 + 2 6 = 11 6
(c)
Consider the given figure (a).
Form the above figure, it is observed that the given circuit are in same pattern and in first loop two elements of 1 Ω resistors are in series and 2 Ω resistors are in series and their combination are in parallel.
The equivalent resistance of each loop is given as,
R = ( R 1 + R 2 ) ⋅ ( R 3 + R 4 ) ( R 1 + R 2 ) + ( R 3 + R 4 )
Where, the combination of ( R 1 + R 2 ) is ( 1 + 1 ) Ω and the combination of ( R 3 + R 4 ) is ( 2 + 2 ) Ω .
By substituting the given values in the above expression, we get
R = ( 1 + 1 ) ⋅ ( 2 + 2 ) ( 1 + 1 ) + ( 2 + 2 ) = 2 × 4 2 + 4 = 8 6 = 4 3 Ω
Now, the reduced figure is given as,
The equivalent resistance of the complete circuit is given as,
R eq = R + R + R + R = 4 R
Where, the equivalent resistance of each loop is R .
By substituting the values in the above equation, we get
R eq = 4 × 4 3 = 16 3 Ω
Thus, the equivalent resistance for the given figure (a) is 16 3 Ω .
(ii)
Consider the given figure (b).
From the given circuit it is clear that the all five resistors of resistance R are connected in series.
The equivalent resistance of the circuit is given as,
R eq = R + R + R + R + R = 5 R
Thus, the equivalent resistance for the given figure (b) is 5 R .
a)
Given: The total number of resistors is
(i)
When all the resistors are connected in series, then the effective resistance is the maximum.
The expression for the maximum effective resistance is given as,
Where, the maximum effective resistance is
(ii)
When all the resistors are connected in parallel, then the effective resistance is the minimum.
The expression for the minimum effective resistance is given as,
Where, the minimum effective resistance is
By dividing the equation (1) by (2), we get
Thus, the ratio of the maximum to the minimum resistance is
b)
Given: The resistances are
(i)
Consider the circuit combination for the required equivalent resistance is given as,
The equivalent resistance of the above circuit is given as,
Where, the equivalent resistance is
By substituting the given values in the above expression, we get
(ii)
Consider the circuit combination for the required equivalent resistance is given as,
The expression for the equivalent resistance of the above circuit is given as,
Where, the equivalent resistance is
By substituting the given values in the above expression, we get
(iii)
Consider the circuit combination for the required equivalent resistance is given as,
The expression for the equivalent resistance of the above circuit is given as,
Where, the equivalent resistance is
By substituting the given values in the above expression, we get
(iv)
Consider the circuit combination for the required equivalent resistance is given as,
The expression for the equivalent resistance of the above circuit is given as,
Where, the equivalent resistance is
By substituting the given values in the above expression, we get
(c)
Consider the given figure (a).
Form the above figure, it is observed that the given circuit are in same pattern and in first loop two elements of
The equivalent resistance of each loop is given as,
Where, the combination of
By substituting the given values in the above expression, we get
Now, the reduced figure is given as,
The equivalent resistance of the complete circuit is given as,
Where, the equivalent resistance of each loop is
By substituting the values in the above equation, we get
Thus, the equivalent resistance for the given figure (a) is
(ii)
Consider the given figure (b).
From the given circuit it is clear that the all five resistors of resistance
The equivalent resistance of the circuit is given as,
Thus, the equivalent resistance for the given figure (b) is
