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In the following network potential at 'O'
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The correct answer is:
\( 4.8 \mathrm{~V} \)
According to Kirchhoff's law, we have
\( I_{1}+I_{2}+I_{3}=0 \)
By Ohm's law, we have \( \frac{8-V_{0}}{2}+\frac{4-V_{0}}{4}+\frac{2-V_{0}}{2}=0 \)
\( \Rightarrow \frac{2\left(8-V_{0}\right)+\left(4-V_{0}\right)+2\left(2-V_{0}\right)}{4}=0 \)
\( \Rightarrow 16-2 V_{0}+4-V_{0}+4-2 V_{0}=0 \)
\( \Rightarrow 24-5 V_{0}=0 \)
\( \Rightarrow V_{0}=\frac{24}{5}=4.8 \mathrm{~V} \)
Therefore, network potential at ' \( \mathrm{O} \) ' is \( 4.8 \mathrm{~V} \)
\( I_{1}+I_{2}+I_{3}=0 \)
By Ohm's law, we have \( \frac{8-V_{0}}{2}+\frac{4-V_{0}}{4}+\frac{2-V_{0}}{2}=0 \)
\( \Rightarrow \frac{2\left(8-V_{0}\right)+\left(4-V_{0}\right)+2\left(2-V_{0}\right)}{4}=0 \)
\( \Rightarrow 16-2 V_{0}+4-V_{0}+4-2 V_{0}=0 \)
\( \Rightarrow 24-5 V_{0}=0 \)
\( \Rightarrow V_{0}=\frac{24}{5}=4.8 \mathrm{~V} \)
Therefore, network potential at ' \( \mathrm{O} \) ' is \( 4.8 \mathrm{~V} \)
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