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......... ADVERTISEMENT ......... ..8.. a.) Find the current through resistor
a) in the figure.
b.) Find the potential difference across resistor a) in the figure.
c.) Find the current through resistor b) in the figure.
d.) Find the potential difference across resistor b) in the figure.
e.) Find the current through resistor c) in the figure.
f.) Find the potential difference across resistor c) in the figure.
g.) Find the current through resistor d) in the figure.
h.) Find the potential difference across resistor d) in the figure.
I can tell that this is a combination circuit where some resistors are in a parallel arrangement (B, C+D) and some are in a series such as C and D.
Is this just a straightforward Ohm’s law problem? Or are Kirchoff’s Rules needed (but there is only one power supply?)
I assume V=IR is viable in this circuit but I’d like to be sure.
The resistors b and c are parallel.1 /Rbcd = 1/5 + 1//5 + 1/5
Rbcd = 1.67 Ohm
R eq = Rbcd + Ra = 6.67 Ohm
I = V / Rq = 10 / 6.67 = 1.5 A ….Ans
b) V = V = RI = 5×1.5 = 7.5 V
c) Current 1.5 A goes into all three parrallels b, C & D resistor.
It can be divided equally between all three, as they all have the same resistance.
I = 1.5/3 = 0.50
d) V = R = 5 x 0.25 = 2.5 Volt
e) Ic = 0.25 A
f) Vc = 2.25 V
g) Id = 0.50
h) Vd = 2.25 V