Question
In: Physics
A 63.0-? resistor is connected in parallel with a 115.0-? resistor. This parallel group is connected...
A 63.0-? resistor is connected in parallel with a 115.0-? resistor. This parallel group is connected in series with a 17.0-? resistor. The total combination is connected across a 15.0-V battery.
(a) Find the current in the 115.0-? resistor.
A
(b) Find the power dissipated in the 115.0-? resistor.
W
Solutions
Expert Solution
similar one but with different values helped for the other guy sorry for the inconvenience any doubts ping me
First, find the equivelent resistance of the 63 and 119 ohm
resistors, or...
63||119 = 41.2 ohms
Now, find the equivelent resistance of this and the 15 ohm
resistor, or...
41.2 + 15 = 56.2 ohms
Next, find the current through from the battery, or ...
I = V/R = 15/56.2
I = 0.267 A
This current flows through the 15 ohm resistor, and a part of it
flows through each of the parallel resistors. To find the current
in the parallel resistor, you use a "Current Divider", or ..
First the current in the 63 ohm resistor, or ...
I(63) = 0.267*[119/(63+119)]
I(63) = =0.1745 Amps
To find the current in the other resistor, you can either set up a
"Current Divider" for that one, or just subtract what you just
calculated from the total. I will use the current divider, or
...
I(119) = 0.267*[63/(63+119)]
I(119) = 0.0925 Amps
genius_generous answered 2 hours agoRelated Solutions
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