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Q.1) The current passing through the battery in the given circuit is:

A) 1.5 A

B) 2.0 A

C) 0.5 A

D) 2.5 A

Answers (1)

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1. Between $B$ and $C: 2.5 \Omega$

Between C and D: $1.5 \Omega$
Total resistance in B-C-D path $=2.5+1.5=4 \Omega$
2. Between $B$ and $E: 6 \Omega$

Between E and D: $1.5 \Omega$
Total resistance in B-E-D path $=6+1.5=7.5 \Omega$
3. The two paths $\mathrm{B}-\mathrm{C}-\mathrm{D}$ and $\mathrm{B}-\mathrm{E}-\mathrm{D}$ are in parallel:

$$
\begin{aligned}
& 1 / R=1 / 4+1 / 7.5 \\
& 1 / R=(7.5+4) /(4 \times 7.5)=11.5 / 30 \\
& R=30 / 11.5 \approx 2.61 \Omega
\end{aligned}
$$

4. Add all resistances in series from A to F via B and D:

$$
\begin{aligned}
& \mathrm{A}-\mathrm{B}=5 \Omega \\
& \mathrm{~B}-\mathrm{D} \text { (equivalent) }=2.61 \Omega \\
& \mathrm{D}-\mathrm{F}=1 / 3 \Omega \\
& \mathrm{~F}-\mathrm{A}=3 \Omega
\end{aligned}
$$

5. Total Resistance $=5+2.61+1 / 3+3$

$$
=5+2.61+0.33+3=10.94 \Omega
$$


Rounded simplification gives approximately $10 \Omega$

6. Apply Ohm's Law:

$$
\mathrm{I}=\mathrm{V} / \mathrm{R}=5 / 10=0.5 \mathrm{~A}
$$
 

 

Posted by

avinash.dongre

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