Question

For the circuit shown in figure, the potential difference between points a and b is :-
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Answer 1

Answer:

the answer will be 430.5 volt

Answer 2

Answer:

Potential difference between points a and b is 4.5V.

Explanation:

Find: the potential difference between points a and b.

When resistors are connected one after each other this is called connecting in series.

If R1  R2 and R3 are resistors connected in series, the equivalent resistor is

$R_{s} =R_{1} +R_{2}+R_{3}$

When resistors are connected across each other (side by side) this is called connecting in parallel.

If R1  R2 and R3 are resistors connected in parallel, the equivalent resistor is:

$\frac{1}{R_{p} }=\frac{1}{R_{1} } +\frac{1}{R_{2} }+\frac{1}{R_{3} }$

Resistor circuits that combine series and parallel resistors networks together are generally known as Resistor Combination or mixed resistor circuits. The method of calculating the circuits equivalent resistance is the same as that for any individual series or parallel circuit.

Resistors in series carry exactly the same current and that resistors in parallel have exactly the same voltage across them.

The equivalent circuit is shown in diagram:

Equivalent resistance of R2,R3 and R4

$\frac{1}{R_{p1} } =\frac{1}{R_{2} } +\frac{1}{R_{3} }+\frac{1}{R_{4} }\\\frac{1}{R_{p1} } =\frac{1}{15} +\frac{1}{30} +\frac{1}{10} \\\frac{1}{R_{p1} } =\frac{2+1+3}{30} \\\frac{1}{R_{p1} } =\frac{6}{30} \\R_{p1} =5\Omega$

Equivalent resistance of R5 and R6

$\frac{1}{R_{p2} } =\frac{1}{R_{5} } +\frac{1}{R_{6} } \\\frac{1}{R_{p2} } =\frac{1}{12} +\frac{1}{4} \\\frac{1}{R_{p2} } =\frac{1+3}{12} \\\frac{1}{R_{p2} } =\frac{4}{12} \\R_{p2}=3\Omega$

Given, potential difference across R6 is 15V which is also the voltage across Rp2.

Current flowing across Rp2 is

$I_{p2}=V_{p2}\times R_{p2}\\I_{p2}=15\times3\\I_{p2}=45A$

Equivalent resistance between points a and b is

$R_{ab}=(1+5+3+1)\Omega\\R_{ab}=10\Omega$

In series, current are same in all resistance, therefore current flowing across points a and b is 45A.

Hence, potential difference between points a and b is

$V_{ab}=\frac{I_{ab} }{R_{ab} } \\V_{ab}=\frac{45}{10}\\ V_{ab}=4.5V$

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