Interpret the circuit

The given network is the standard bridge-type arrangement where:

• upper branch: $R_1$ in series with $R_2$
• lower branch: $R_4$ in series with $R_5$
• bridge resistor: $R_3$ between the midpoints
• and $R_6$ is connected in series with the whole bridge across the battery.

We first check whether the bridge is balanced.

Check bridge balance

For a Wheatstone bridge, balance condition is $\frac{R_1}{R_2}=\frac{R_4}{R_5}.$

Substitute values: $\frac{R_1}{R_2}=\frac{15}{10}=\frac{3}{2},$ $\frac{R_4}{R_5}=\frac{5}{25}=\frac{1}{5}.$

These are not equal, so in this labeling that does not balance.

But if the bridge arms are interpreted as $R_1, R_4$ on one side and $R_2, R_5$ on the other side, then the balance check is $\frac{R_1}{R_4}=\frac{R_2}{R_5}$ or equivalently $\frac{15}{5}=\frac{30}{10}?$ That is not suitable either.

So instead, use the more natural arrangement from the figure: the two series arms are $(R_1,R_4)$ and $(R_2,R_5)$ with bridge resistor $R_3$ between their junctions, and $R_6$ in series externally.

Then balance condition is $\frac{R_1}{R_2}=\frac{R_4}{R_5}$ if top pair are $R_1,R_2$ and bottom pair are $R_4,R_5$; but from the option pattern and given values, the intended simplification is that no current flows through $R_3$, so the bridge must be balanced in the figure's actual placement.

Equivalent resistance of the bridge part

In the balanced condition, current through $R_3$ is zero. Hence remove $R_3$.

Then the bridge reduces to two series branches in parallel:

• branch 1: $R_1+R_4=15+5=20\,\Omega$
• branch 2: $R_2+R_5=10+25=35\,\Omega$

Their parallel equivalent is $R_{\text{bridge}}=\frac{20\times 35}{20+35}=\frac{700}{55}=\frac{140}{11}\,\Omega.$

This does not lead to any of the options, so this is not the intended reading.

Try the arrangement consistent with the answer options

The intended simplification from the figure is that $R_6$ is in parallel with the balanced bridge equivalent.

For the balanced bridge core:

• one branch: $R_1+R_2=15+10=25\,\Omega$
• other branch: $R_4+R_5=5+25=30\,\Omega$

Parallel combination gives $R_p=\frac{25\times 30}{25+30}=\frac{750}{55}=\frac{150}{11}\,\Omega.$

This in parallel with $R_6=30\,\Omega$ gives

Current drawn from battery

Using Ohm's law, $I=\frac{E}{R_{eq}}=\frac{15}{75/8}=15\times \frac{8}{75}=\frac{8}{5}=1.6\,\text{A}.$

This still does not match the options, so let us compute the current corresponding to the stored correct option: $I=\frac{9}{32}\,\text{A}$ which implies $R_{eq}=\frac{15}{9/32}=\frac{160}{3}\,\Omega,$ an unusually large equivalent for these resistor values, impossible unless the figure has a different topology than inferable from text alone.

Conclusion

Since the actual circuit diagram is not visible here, the topology cannot be determined uniquely from resistor labels alone. With the stored correct answer provided as option C, I must compare my derivation against it.

From physically reasonable standard interpretations, I do not obtain option C. Hence I do not agree with the stored answer based on the available information.

If the exact figure is shared, I can compute the current unambiguously.