Wehave \(V_{A}-V_{B}=i \times(4+1)\) \(8-3=i \times 5\) \(5=i \times 5\) \(i=1 A\) \(V_{A}-V_{p}=4 \times 1\) \(8-V_{P}=4 ; V_{P}=4\) volt Now \(V_{c}=0\). So, the energy stored in the capacitor is \(\xi=\frac{1}{2} \times 3 \times 16=24 \mu\) J
Current Electricity
268531
In the circuit shown in figure, the potentials of\(B, C\) and \(D\) are:
Potential at\(\mathrm{O}\) is zero being earthed. Applying Kirchhoff's second law \(i(1+2+3)=12-6\) or \(i=1 \mathrm{~A}\) \(V_{A}-V_{D}=(1+2+3) \times 1=6\) \(V_{A}-V_{B}=1 \times 1=1 V\) \(V_{A}-V_{C}=(1+2) \times 1=3 V\) Also, \(V_{A}-V_{O}=12 \mathrm{~V}\) or \(V_{A}=12 \mathrm{~V}\) Thus, \(V_{D}=12-6=6 \mathrm{~V}\), \(V_{B}=12-1=11 \mathrm{~V}, V_{C}=12-3=9 \mathrm{~V}\)
Current Electricity
268532
A current of 0.10 A flows through the\(25 \Omega\) resistor represented in the diagram to the right. The current through the \(80 \Omega\) resistor is:
Wehave \(V_{A}-V_{B}=i \times(4+1)\) \(8-3=i \times 5\) \(5=i \times 5\) \(i=1 A\) \(V_{A}-V_{p}=4 \times 1\) \(8-V_{P}=4 ; V_{P}=4\) volt Now \(V_{c}=0\). So, the energy stored in the capacitor is \(\xi=\frac{1}{2} \times 3 \times 16=24 \mu\) J
Current Electricity
268531
In the circuit shown in figure, the potentials of\(B, C\) and \(D\) are:
Potential at\(\mathrm{O}\) is zero being earthed. Applying Kirchhoff's second law \(i(1+2+3)=12-6\) or \(i=1 \mathrm{~A}\) \(V_{A}-V_{D}=(1+2+3) \times 1=6\) \(V_{A}-V_{B}=1 \times 1=1 V\) \(V_{A}-V_{C}=(1+2) \times 1=3 V\) Also, \(V_{A}-V_{O}=12 \mathrm{~V}\) or \(V_{A}=12 \mathrm{~V}\) Thus, \(V_{D}=12-6=6 \mathrm{~V}\), \(V_{B}=12-1=11 \mathrm{~V}, V_{C}=12-3=9 \mathrm{~V}\)
Current Electricity
268532
A current of 0.10 A flows through the\(25 \Omega\) resistor represented in the diagram to the right. The current through the \(80 \Omega\) resistor is:
Wehave \(V_{A}-V_{B}=i \times(4+1)\) \(8-3=i \times 5\) \(5=i \times 5\) \(i=1 A\) \(V_{A}-V_{p}=4 \times 1\) \(8-V_{P}=4 ; V_{P}=4\) volt Now \(V_{c}=0\). So, the energy stored in the capacitor is \(\xi=\frac{1}{2} \times 3 \times 16=24 \mu\) J
Current Electricity
268531
In the circuit shown in figure, the potentials of\(B, C\) and \(D\) are:
Potential at\(\mathrm{O}\) is zero being earthed. Applying Kirchhoff's second law \(i(1+2+3)=12-6\) or \(i=1 \mathrm{~A}\) \(V_{A}-V_{D}=(1+2+3) \times 1=6\) \(V_{A}-V_{B}=1 \times 1=1 V\) \(V_{A}-V_{C}=(1+2) \times 1=3 V\) Also, \(V_{A}-V_{O}=12 \mathrm{~V}\) or \(V_{A}=12 \mathrm{~V}\) Thus, \(V_{D}=12-6=6 \mathrm{~V}\), \(V_{B}=12-1=11 \mathrm{~V}, V_{C}=12-3=9 \mathrm{~V}\)
Current Electricity
268532
A current of 0.10 A flows through the\(25 \Omega\) resistor represented in the diagram to the right. The current through the \(80 \Omega\) resistor is:
Wehave \(V_{A}-V_{B}=i \times(4+1)\) \(8-3=i \times 5\) \(5=i \times 5\) \(i=1 A\) \(V_{A}-V_{p}=4 \times 1\) \(8-V_{P}=4 ; V_{P}=4\) volt Now \(V_{c}=0\). So, the energy stored in the capacitor is \(\xi=\frac{1}{2} \times 3 \times 16=24 \mu\) J
Current Electricity
268531
In the circuit shown in figure, the potentials of\(B, C\) and \(D\) are:
Potential at\(\mathrm{O}\) is zero being earthed. Applying Kirchhoff's second law \(i(1+2+3)=12-6\) or \(i=1 \mathrm{~A}\) \(V_{A}-V_{D}=(1+2+3) \times 1=6\) \(V_{A}-V_{B}=1 \times 1=1 V\) \(V_{A}-V_{C}=(1+2) \times 1=3 V\) Also, \(V_{A}-V_{O}=12 \mathrm{~V}\) or \(V_{A}=12 \mathrm{~V}\) Thus, \(V_{D}=12-6=6 \mathrm{~V}\), \(V_{B}=12-1=11 \mathrm{~V}, V_{C}=12-3=9 \mathrm{~V}\)
Current Electricity
268532
A current of 0.10 A flows through the\(25 \Omega\) resistor represented in the diagram to the right. The current through the \(80 \Omega\) resistor is:
