Five 5.0 Ohm resistors are connected in series with a 21.0 V battery. Find the f

Question

Five 5.0 Ohm resistors are connected in series with a 21.0 V battery. Find the following. (a) the equivalent resistance of the circuit Ohm (b) the current in each resistor A (c) Repeat for the case in which all five resistors are connected in parallel across the battery. equivalent resistance Ohm current in each resistor A Consider the following figure. (a) Find the equivalent resistance between points a and b in the figure. (R = 12.0 Ohm) Ohm (b) Calculate the current in each resistor if a potential difference of 32.0 V is applied between points a and b. I (4.00 Ohm) = A I (7.00 Ohm) = A I (12.0 Ohm) = A I (9.00 Ohm) = A

Explanation / Answer

1) r tot = 5*r = 5*5 = 25 ohm

b) form ohms law v = ir

i = v/r = 21/25 = 0.84 amp

c) in parallel rtot = r/5 = 5/5 = 1 ohm

voltage across each resistor is same 21 V

so i = v/r = 21/1 = 21 amp

2) a) 7 and 12 ohm are in parallel = 7*12/19 = 4.42 ohm

4.42,4 and 9 ohm are in series

r tot = 4.42+4+9 = 17.42 ohm

b) i tot = v/r = 32/17.42 = 1.84 amp

in series i same i4= i9 = 1.84 amp

i 4.42 = 1.84 amp

v 4.42 = 1.84*4.42 = 8.13 V

in parallel voltage same

v7= v12 = 8.13V

i7 = 8.13/7 = 1.16 amp

i12 = 8.13/12 = 0.68 amp

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