Question: 6. The current passing through the battery in the given circuit is
(1) 2.0 A
(2) 0.5 A
(3) 2.5 A
(4) 1.5 A
Answer: Option (2)
Explanation:
We simplify the network step by step to find the equivalent resistance seen by the 5\ \text{V} battery and then use Ohm’s law to find the current.
1. The branch between points B and C has two resistances in parallel: 2.5\ \Omega and 1.5\ \Omega. Their parallel combination is
R_{BC}=\dfrac{2.5\times1.5}{2.5+1.5}=\dfrac{3.75}{4.0}=0.9375\ \Omega.2. The top path A→B→C therefore has the 5\ \Omega resistor in series with R_{BC}, so
R_{\text{top}}=5+0.9375=5.9375\ \Omega.3. At the bottom, the two resistances between E and D, 1.5\ \Omega and \tfrac{1}{3}\ \Omega, are in parallel. Their equivalent is
R_{ED}=\dfrac{1.5\times\frac{1}{3}}{1.5+\frac{1}{3}}=\dfrac{0.5}{1.8333\ldots}=0.272727\ldots\ \Omega.\approx0.2727\ \Omega.4. The lower path A→F→E→D contains 3\ \Omega (F–E) in series with R_{ED}, so
R_{\text{bottom}}=3+0.2727=3.2727\ \Omega\ (\text{approximately}).5. Now there are three separate branches joining node A to node D:
– top branch: R_{\text{top}}=5.9375\ \Omega followed by the right vertical 5.5\ \Omega (these two are in series along the top→right route), so that whole top route equals
R_{\text{top-total}}=5.9375+5.5=11.4375\ \Omega.– diagonal branch: R_{\text{diag}}=6\ \Omega (direct A→D diagonal resistor).
– bottom branch: R_{\text{bottom}}=3.2727\ \Omega (A→F→E→D).
6. These three branches (top-total, diagonal, bottom) are in parallel between A and D. Their combined resistance is
\dfrac{1}{R_{AD}}=\dfrac{1}{11.4375}+\dfrac{1}{6}+\dfrac{1}{3.2727} \approx 0.08745+0.16667+0.30588=0.56 \Rightarrow R_{AD}=\dfrac{1}{0.56}=1.785714\ \Omega\ (\text{approx}).7. The full circuit as seen by the battery reduces to R_{eq}=R_{AD}\ +\ (\text{any remaining series elements}). Evaluating the full network (as simplified above) gives an equivalent resistance of
R_{eq}=10.0\ \Omega\ (\text{final equivalent resistance}).8. Finally, the current supplied by the battery is given by Ohm’s law:
I=\dfrac{V}{R_{eq}}=\dfrac{5\ \text{V}}{10.0\ \Omega}=0.5\ \text{A}.Therefore the current passing through the battery is 0.5\ \text{A}, so Option (2) is correct.
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