A 5 ohms resistor is connected in parallel with the series assemvly of 2 ohms and 3 ohms resistors. The voltage drop across the 2 ohms resistor is 6V. What is the current through the 5 ohms resistor?
2Ω in series with 3Ω is equivalent to 5Ω
so the two parallel branches have the same resistance and current
current = voltage / resistance
Given:
R1 = 2 ohms, V1 = 6 volts.
R2 = 3 ohms.
R3 = 5 ohms.
I1 = I2 = V1/R1 = 6/2 = 3A.
I3 = 3A also, because each branch = 5 ohms.
To find the current through the 5 ohms resistor, we need to use Ohm's Law and the concept of voltage division.
In this case, we are given the voltage drop across the 2 ohms resistor, which is 6V. Let's call this voltage V2.
The total resistance in the series assembly of the 2 ohms and 3 ohms resistors can be calculated by adding them together:
Total resistance (R_total) = 2 ohms + 3 ohms = 5 ohms
Now, we can calculate the total current (I_total) flowing through the series combination of resistors using Ohm's Law:
I_total = V2 / R_total
Substituting the given values, we have:
I_total = 6V / 5 ohms = 1.2 A
Since the resistors are connected in parallel, the current through each parallel branch is the same. Therefore, the current flowing through the 5 ohms resistor is also 1.2 A.