Three resistors having resistances of 1-ohm, 2-ohm and 3-ohm ,are connected in series to a 12-volt
battery having negligible internal resistance.
connected across a 12-volt source line.
battery having negligible internal resistance.
When resistors are connected in parallel, the total resistance (Rt) is given by the formula:
1/Rt = 1/R1 + 1/R2 + 1/R3 + ...
In this case, we have three resistors connected in parallel: a 60-ohm resistor (R1), a 90-ohm resistor (R2), and a 120-ohm resistor (R3).
Let's calculate the equivalent resistance:
1/Rt = 1/60 + 1/90 + 1/120
To add fractions, we need to find a common denominator. In this case, the common denominator is 360:
1/Rt = (6/360) + (4/360) + (3/360)
1/Rt = 13/360
To find Rt, we take the reciprocal of both sides:
Rt = 360/13 ≈ 27.69 ohms
Therefore, the equivalent resistance of the parallel combination is approximately 27.69 ohms.
Now, let's calculate the total current (It) flowing through this combination using Ohm's Law:
It = V/Rt
Here, V represents the voltage across the combination, which is 12 volts.
It = 12 V / 27.69 Ω ≈ 0.433 A
So, the total current flowing through the parallel combination is approximately 0.433 Amperes.