Question

Q8. For below circuit diagram find-
i. ammeter reading
ii. Equivalent resistance


Given: R1 = 10 ohm, R2 = 40 ohm, R3 = 30 ohm R, = 20 ohm. Rs = 60 ohm

Answer 1

Given

A circuit with R₁ = 10 Ω , R₂ = 40 Ω , R₃ = 30 Ω , R₄ = 20 Ω , R₅ = 60 Ω

To Find

i. Ammeter reading ( Current )

ii. Equivalent resistance

Knowledge Required

If n resistors are connected in series ,

$\bf \pink{\bigstar\ \; R_{eq}=R_1+R_2+R_3+...+R_n}$

If n resistors are connected in series ,

$\bf \green{\bigstar\ \; \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...+\dfrac{1}{R_n}}$

Solution

R₁ , R₂ are connected in series ,

$\to \rm \dfrac{1}{R_{12}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\to \rm \dfrac{1}{R_{12}}=\dfrac{1}{10}+\dfrac{1}{40}\\\\\to \rm \dfrac{1}{R_{12}}=\dfrac{5}{40}\\\\\to \rm \dfrac{1}{R_{12}}=\dfrac{1}{8}\\\\\to \bf R_{12}=8\ \Omega\ \; \bigstar$

R₃ , R₄ , R₅ are too connected in parallel ,

$\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{60}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{20}+\dfrac{3}{60}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{20}+\dfrac{1}{20}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{2}{20}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{10}\\\\\to \bf R_{345}=10\ \Omega\ \; \bigstar$

Now , R₁₂ and R₃₄₅ are connected in series ,

$\to \rm R_{12345}=R_{12}+R_{345}\\\\\to \rm R_{12345}=8+10\\\\\to \rm R_{12345}=18\ \Omega\\\\\bf \to \blue{R_{eq}=18\ \Omega\ \; \bigstar}$

ii. Equivalent resistance = 18 Ω

Apply Ohms Law for finding current in the circuit ,

$\to \bf V=IR_{eq}\\\\\to \rm 12=I(18)\\\\\to \rm I=\dfrac{2}{3}\\\\\to \rm I=0.666\ A\\\\\bf \to \red{I\cong 0.67\ A\ \; \bigstar}$

i. Ammeter reading = 0.67 A

Answer 2

Answer:

Given

A circuit with R₁ = 10 Ω , R₂ = 40 Ω , R₃ = 30 Ω , R₄ = 20 Ω , R₅ = 60 Ω

To Find

i. Ammeter reading ( Current )

ii. Equivalent resistance

Knowledge Required

If n resistors are connected in series ,

\bf \pink{\bigstar\ \; R_{eq}=R_1+R_2+R_3+...+R_n}★ R

eq

=R

1

+R

2

+R

3

+...+R

n

If n resistors are connected in series ,

\bf \green{\bigstar\ \; \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...+\dfrac{1}{R_n}}★

R

eq

1

=

R

1

1

+

R

2

1

+

R

3

1

+...+

R

n

1

Solution

R₁ , R₂ are connected in series ,

\begin{gathered}\to \rm \dfrac{1}{R_{12}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\to \rm \dfrac{1}{R_{12}}=\dfrac{1}{10}+\dfrac{1}{40}\\\\\to \rm \dfrac{1}{R_{12}}=\dfrac{5}{40}\\\\\to \rm \dfrac{1}{R_{12}}=\dfrac{1}{8}\\\\\to \bf R_{12}=8\ \text{\O}mega\ \; \bigstar\end{gathered}

→

R

12

1

=

R

1

1

+

R

2

1

→

R

12

1

=

10

1

+

40

1

→

R

12

1

=

40

5

→

R

12

1

=

8

1

→R

12

=8 Ømega ★

R₃ , R₄ , R₅ are too connected in parallel ,

\begin{gathered}\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{60}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{20}+\dfrac{3}{60}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{20}+\dfrac{1}{20}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{2}{20}\\\\\to \rm \dfrac{1}{R_{345}}=\dfrac{1}{10}\\\\\to \bf R_{345}=10\ \text{\O}mega\ \; \bigstar\end{gathered}

→

R

345

1

=

R

3

1

+

R

4

1

+

R

5

1

→

R

345

1

=

30

1

+

20

1

+

60

1

→

R

345

1

=

20

1

+

60

3

→

R

345

1

=

20

1

+

20

1

→

R

345

1

=

20

2

→

R

345

1

=

10

1

→R

345

=10 Ømega ★

Now , R₁₂ and R₃₄₅ are connected in series ,

\begin{gathered}\to \rm R_{12345}=R_{12}+R_{345}\\\\\to \rm R_{12345}=8+10\\\\\to \rm R_{12345}=18\ \text{\O}mega\\\\\bf \to \blue{R_{eq}=18\ \text{\O}mega\ \; \bigstar}\end{gathered}

→R

12345

=R

12

+R

345

→R

12345

=8+10

→R

12345

=18 Ømega

→R

eq

=18 Ømega ★

ii. Equivalent resistance = 18 Ω

Apply Ohms Law for finding current in the circuit ,

\begin{gathered}\to \bf V=IR_{eq}\\\\\to \rm 12=I(18)\\\\\to \rm I=\dfrac{2}{3}\\\\\to \rm I=0.666\ A\\\\\bf \to \red{I\cong 0.67\ A\ \; \bigstar}\end{gathered}

→V=IR

eq

→12=I(18)

→I=

3

2

→I=0.666 A

→I≅0.67 A ★

i. Ammeter reading = 0.67 A