Question

In the given circuit, find : (A) Total resistance of the network of resistors (B) Current through ammeter A, and
(C) Potential difference across 3ohm and 6ohm resistors

Answer 1

Answer:

answer is C

Explanation:

(C)potential difference across 30hm and 60hm resistors

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Answer 2

The circuit is attached below.

Given:

The circuit is attached below.

Formula Used:

Equivalent resistance of resistors connected in series:

$R=R_1+R_2+R_3+...$

Equivalent resistance of resistors connected in parallel:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...$

Ohm's law:

$V=IR$

where, V is the potential difference, I is the current and R is the resistance.

Calculations:

(A)

Resistors $4\Omega$ and $2\Omega$ are connected in series. $R=4+2 = 6\Omega$

Resistors $3\Omega$ and

R and R' are connected in parallel. $I'=\frac{V}{R'}=\frac{6V}{6\Omega}=1 A$

$R" = (\frac{1}{6}+\frac{1}{6})^{-1} =3\Omega$

(B)

Potential difference, V= 6V

From Ohm's law,

$V=IR\\I=\frac{V}{R}\\I=\frac{6V}{3\Omega}=2A$

(C) Potential difference across each branch in parallel connection is same.

So, the branch containing resistor $3\Omega$ has potential difference 6 V.

Current in the branch would be, $I'=\frac{V}{R'}=\frac{6V}{6\Omega}=1 A$

Potential difference across $3\Omega$ would be $V_3=(1A)(3\Omega)=3V$

Learn more about: Ohm's law and equivalent resistance

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