Question

- r.2 i A, i(10 Ω) = 1.03 A, and i(962) = 2.1 A. c) I(4 Ω) = 2.3 A, l(7 Ω) = 1 .644 A, l(10 Ω) = 1.23 A, and i(9 Ω) = 2.3 A. D) l(4 Ω) = 2.6A, l(7 Ω) = 1 .92 A, l(10 Ω) = 1.63 A, and I(96) = 2.6 A. 6) Consider the combination of resistors shown. Find the total equivalent resistance between the points (a) and (b) 12.0 Ω 4.00 Ω 5.00 Ω 6.00 Ω 8.00Ω A)4.12Ω; B)8.15Ω; a 1 1.67 Ω; or D) 23.22 Ω. 7) For the circuit in Problem-6, also calculate the current flowing through resistor, if a 34 V potential difference is applied across the points (a) and A) 1(12 Ω) = 0.33 A, I(6 Ω) = 0.67 A, l(5 Ω) = 1.00 A, l(8 Ω) = 0.33 A, 1(4 Ω 0.67 A B) 1(12 Ohm) = 0.66 A, l(6 Ω) = 1.34 A, l(5 Ω) = 2.00 A, l(8 Ω) = 0.66 A, I(

Answer 1

Your answer to problem 6 is correct.

Problem 7:

Total current in the circuit is

$i=\frac{V}{R}=\frac{34}{11.67}A=2.913A$

Now, as 5 ohm is connected in series, current through it is 2.913 A

For combination of 12 ohm and 6 ohm in parallel, their net resistance is 4 ohm. Thus, voltage across them is 4*2.913 A = 11.652 volt

$i(6\Omega )=\frac{11.652}{6}=1.942A$

$i(12\Omega )=\frac{11.652}{12}=0.971A$

Similarly, net resistance of 4 ohm and 8 ohm resistors in parallel is 2.67 ohm.

Voltage across them is

V 2.913 * 2.67uolt-7.77 uolts

Therefore,

$i(4\Omega )=\frac{7.77}{4}=1.9425A$ and

$i(8\Omega )=\frac{7.77}{8}=0.97125A$