Solveeit Logo

Question

Question: If the X = 12 V, Y = 33 $\Omega$, the current IT is _______ A....

If the X = 12 V, Y = 33 Ω\Omega, the current IT is _______ A.

Answer

0.3367

Explanation

Solution

To find the total current ITI_T in the circuit, we need to calculate the equivalent resistance (ReqR_{eq}) of the entire circuit and then apply Ohm's Law (IT=X/ReqI_T = X / R_{eq}).

Given values: Voltage X=12X = 12 V Resistance Y=33ΩY = 33 \, \Omega

Let's break down the circuit to find the equivalent resistance:

  1. Identify the series combination of the 3 Ω\Omega and 6 Ω\Omega resistors:
    The diagram shows that the current flowing through the 3 Ω\Omega resistor then flows through the 6 Ω\Omega resistor before reaching the common bottom line. Thus, they are in series.
    Rs1=3Ω+6Ω=9ΩR_{s1} = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega.

  2. Identify the parallel combination of the 2 Ω\Omega resistor and Rs1R_{s1}:
    The 2 Ω\Omega resistor is connected in parallel with the series combination Rs1R_{s1} (9 Ω\Omega). Both branches start at the same node (after the 1 Ω\Omega resistor) and end at the common bottom line.
    Rp1=2Ω×Rs12Ω+Rs1=2×92+9=1811ΩR_{p1} = \frac{2 \, \Omega \times R_{s1}}{2 \, \Omega + R_{s1}} = \frac{2 \times 9}{2 + 9} = \frac{18}{11} \, \Omega.

  3. Calculate the total equivalent resistance (ReqR_{eq}):
    The 1 Ω\Omega resistor, the parallel combination Rp1R_{p1} (18/11 Ω\Omega), and the Y Ω\Omega resistor are all connected in series with the voltage source.
    Req=1Ω+Rp1+YΩR_{eq} = 1 \, \Omega + R_{p1} + Y \, \Omega.
    Substitute the given value Y=33ΩY = 33 \, \Omega:
    Req=1+1811+33=34+1811ΩR_{eq} = 1 + \frac{18}{11} + 33 = 34 + \frac{18}{11} \, \Omega.
    To add these, find a common denominator:
    Req=34×1111+1811=374+1811=39211ΩR_{eq} = \frac{34 \times 11}{11} + \frac{18}{11} = \frac{374 + 18}{11} = \frac{392}{11} \, \Omega.

  4. Calculate the total current (ITI_T) using Ohm's Law:
    IT=XReqI_T = \frac{X}{R_{eq}}.
    Substitute the given voltage X=12X = 12 V and the calculated ReqR_{eq}:
    IT=12 V39211Ω=12×11392=132392I_T = \frac{12 \text{ V}}{\frac{392}{11} \, \Omega} = \frac{12 \times 11}{392} = \frac{132}{392} A.

  5. Simplify the fraction:
    Divide both the numerator and the denominator by their greatest common divisor, which is 4:
    IT=132÷4392÷4=3398I_T = \frac{132 \div 4}{392 \div 4} = \frac{33}{98} A.

The decimal value of 3398\frac{33}{98} is approximately 0.33670.3367.