NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
268533
In Wheat stone's bridge shown in the adjoining figure galvanometer gives no deflection on pressing the key, the balance condition for the bridgeis :
At balance, the potentials of point\(B\) and \(D\) are same and there will be no current in the arm BD. Thus, \(\begin{equation*} i_{1} R_{1}=\frac{q}{C_{1}} \tag{i} \end{equation*}\) where \(\mathrm{q}\) is the charge on both the capacitor plates connected in series. Q(uite similarly \(V_{B}-V_{C}=V_{D}-V_{C}\) or \(i_{1} R_{2}=\frac{q}{C_{2}}\) Dividing eqs. (i) and (ii), we get \(\frac{R_{1}}{R_{2}}=\frac{C_{2}}{C_{1}}\)
Current Electricity
268534
In the steady state, the energy stored in the capacitoris :
Whenthecapacitor plateacquirefull cahrge \(q_{0}\), there will be no current in the capacitor arm Applying Kirchhoff's second law to the current carrying circuit \(i\left(R_{1}+R_{2}\right)=E_{1}-i r_{1}\) or \(\quad i=\frac{E_{1}}{r_{1}+R_{1}+R_{2}}\) Now \(V_{a}-V_{b}=-i R_{1}=-\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\) and \(V_{C}=\frac{q_{0}}{C}=E_{2}+i R_{1}=E_{2}+\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\) Now energy stored in the capacitor \(U=\frac{1}{2} C V_{c}^{2} ;=\frac{1}{2} C\left[E_{2}+\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\right]^{2}\)
Current Electricity
268535
A part of circuit in steady state along with the currents flowing in the branches, the value of resistances is shown in figure. Calculate the energy stored in the capacitor.
1 \(8 \times 10^{-1} \mathrm{~J}\)
2 \(8 \times 10^{-2} \mathrm{~J}\)
3 \(8 \times 10^{-3} \mathrm{~J}\)
4 \(8 \times 10^{-4} \mathrm{~J}\)
Explanation:
Whenthecapacitor plates get fully charged, there will be no current in branch ab, Remember capacitance acts as the open circuit since capacitance offers infinite resistance to d.c.The capaditance simply collects the charge. Applying Kirchooff's first law to the jundions a and b, we find \(i_{1}=3 A\) and \(i_{2}=1 A\). Now applying Kirchhoff's second law to the closed mesh aefba, weget \(3 \times 5+3 \times 1+1 \times 2=V_{a}-V_{b}\) \(V_{a}-V_{b}=20 \mathrm{~V}\) Energy stored in the capacitor \(U=\frac{1}{2} C\left(V_{a}-V_{b}\right)^{2}=\frac{1}{2} \times 4 \times 10^{-6} \times(20)^{2}\) \(=8 \times 10^{-4} \mathrm{~J}\)
Current Electricity
268536
Equivalent resistance across\(A\) and \(B\) in the given circuit if \(r=10_{\Omega}, R=20 \Omega\) is
268533
In Wheat stone's bridge shown in the adjoining figure galvanometer gives no deflection on pressing the key, the balance condition for the bridgeis :
At balance, the potentials of point\(B\) and \(D\) are same and there will be no current in the arm BD. Thus, \(\begin{equation*} i_{1} R_{1}=\frac{q}{C_{1}} \tag{i} \end{equation*}\) where \(\mathrm{q}\) is the charge on both the capacitor plates connected in series. Q(uite similarly \(V_{B}-V_{C}=V_{D}-V_{C}\) or \(i_{1} R_{2}=\frac{q}{C_{2}}\) Dividing eqs. (i) and (ii), we get \(\frac{R_{1}}{R_{2}}=\frac{C_{2}}{C_{1}}\)
Current Electricity
268534
In the steady state, the energy stored in the capacitoris :
Whenthecapacitor plateacquirefull cahrge \(q_{0}\), there will be no current in the capacitor arm Applying Kirchhoff's second law to the current carrying circuit \(i\left(R_{1}+R_{2}\right)=E_{1}-i r_{1}\) or \(\quad i=\frac{E_{1}}{r_{1}+R_{1}+R_{2}}\) Now \(V_{a}-V_{b}=-i R_{1}=-\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\) and \(V_{C}=\frac{q_{0}}{C}=E_{2}+i R_{1}=E_{2}+\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\) Now energy stored in the capacitor \(U=\frac{1}{2} C V_{c}^{2} ;=\frac{1}{2} C\left[E_{2}+\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\right]^{2}\)
Current Electricity
268535
A part of circuit in steady state along with the currents flowing in the branches, the value of resistances is shown in figure. Calculate the energy stored in the capacitor.
1 \(8 \times 10^{-1} \mathrm{~J}\)
2 \(8 \times 10^{-2} \mathrm{~J}\)
3 \(8 \times 10^{-3} \mathrm{~J}\)
4 \(8 \times 10^{-4} \mathrm{~J}\)
Explanation:
Whenthecapacitor plates get fully charged, there will be no current in branch ab, Remember capacitance acts as the open circuit since capacitance offers infinite resistance to d.c.The capaditance simply collects the charge. Applying Kirchooff's first law to the jundions a and b, we find \(i_{1}=3 A\) and \(i_{2}=1 A\). Now applying Kirchhoff's second law to the closed mesh aefba, weget \(3 \times 5+3 \times 1+1 \times 2=V_{a}-V_{b}\) \(V_{a}-V_{b}=20 \mathrm{~V}\) Energy stored in the capacitor \(U=\frac{1}{2} C\left(V_{a}-V_{b}\right)^{2}=\frac{1}{2} \times 4 \times 10^{-6} \times(20)^{2}\) \(=8 \times 10^{-4} \mathrm{~J}\)
Current Electricity
268536
Equivalent resistance across\(A\) and \(B\) in the given circuit if \(r=10_{\Omega}, R=20 \Omega\) is
268533
In Wheat stone's bridge shown in the adjoining figure galvanometer gives no deflection on pressing the key, the balance condition for the bridgeis :
At balance, the potentials of point\(B\) and \(D\) are same and there will be no current in the arm BD. Thus, \(\begin{equation*} i_{1} R_{1}=\frac{q}{C_{1}} \tag{i} \end{equation*}\) where \(\mathrm{q}\) is the charge on both the capacitor plates connected in series. Q(uite similarly \(V_{B}-V_{C}=V_{D}-V_{C}\) or \(i_{1} R_{2}=\frac{q}{C_{2}}\) Dividing eqs. (i) and (ii), we get \(\frac{R_{1}}{R_{2}}=\frac{C_{2}}{C_{1}}\)
Current Electricity
268534
In the steady state, the energy stored in the capacitoris :
Whenthecapacitor plateacquirefull cahrge \(q_{0}\), there will be no current in the capacitor arm Applying Kirchhoff's second law to the current carrying circuit \(i\left(R_{1}+R_{2}\right)=E_{1}-i r_{1}\) or \(\quad i=\frac{E_{1}}{r_{1}+R_{1}+R_{2}}\) Now \(V_{a}-V_{b}=-i R_{1}=-\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\) and \(V_{C}=\frac{q_{0}}{C}=E_{2}+i R_{1}=E_{2}+\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\) Now energy stored in the capacitor \(U=\frac{1}{2} C V_{c}^{2} ;=\frac{1}{2} C\left[E_{2}+\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\right]^{2}\)
Current Electricity
268535
A part of circuit in steady state along with the currents flowing in the branches, the value of resistances is shown in figure. Calculate the energy stored in the capacitor.
1 \(8 \times 10^{-1} \mathrm{~J}\)
2 \(8 \times 10^{-2} \mathrm{~J}\)
3 \(8 \times 10^{-3} \mathrm{~J}\)
4 \(8 \times 10^{-4} \mathrm{~J}\)
Explanation:
Whenthecapacitor plates get fully charged, there will be no current in branch ab, Remember capacitance acts as the open circuit since capacitance offers infinite resistance to d.c.The capaditance simply collects the charge. Applying Kirchooff's first law to the jundions a and b, we find \(i_{1}=3 A\) and \(i_{2}=1 A\). Now applying Kirchhoff's second law to the closed mesh aefba, weget \(3 \times 5+3 \times 1+1 \times 2=V_{a}-V_{b}\) \(V_{a}-V_{b}=20 \mathrm{~V}\) Energy stored in the capacitor \(U=\frac{1}{2} C\left(V_{a}-V_{b}\right)^{2}=\frac{1}{2} \times 4 \times 10^{-6} \times(20)^{2}\) \(=8 \times 10^{-4} \mathrm{~J}\)
Current Electricity
268536
Equivalent resistance across\(A\) and \(B\) in the given circuit if \(r=10_{\Omega}, R=20 \Omega\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
268533
In Wheat stone's bridge shown in the adjoining figure galvanometer gives no deflection on pressing the key, the balance condition for the bridgeis :
At balance, the potentials of point\(B\) and \(D\) are same and there will be no current in the arm BD. Thus, \(\begin{equation*} i_{1} R_{1}=\frac{q}{C_{1}} \tag{i} \end{equation*}\) where \(\mathrm{q}\) is the charge on both the capacitor plates connected in series. Q(uite similarly \(V_{B}-V_{C}=V_{D}-V_{C}\) or \(i_{1} R_{2}=\frac{q}{C_{2}}\) Dividing eqs. (i) and (ii), we get \(\frac{R_{1}}{R_{2}}=\frac{C_{2}}{C_{1}}\)
Current Electricity
268534
In the steady state, the energy stored in the capacitoris :
Whenthecapacitor plateacquirefull cahrge \(q_{0}\), there will be no current in the capacitor arm Applying Kirchhoff's second law to the current carrying circuit \(i\left(R_{1}+R_{2}\right)=E_{1}-i r_{1}\) or \(\quad i=\frac{E_{1}}{r_{1}+R_{1}+R_{2}}\) Now \(V_{a}-V_{b}=-i R_{1}=-\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\) and \(V_{C}=\frac{q_{0}}{C}=E_{2}+i R_{1}=E_{2}+\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\) Now energy stored in the capacitor \(U=\frac{1}{2} C V_{c}^{2} ;=\frac{1}{2} C\left[E_{2}+\frac{E_{1} R_{1}}{r_{1}+R_{1}+R_{2}}\right]^{2}\)
Current Electricity
268535
A part of circuit in steady state along with the currents flowing in the branches, the value of resistances is shown in figure. Calculate the energy stored in the capacitor.
1 \(8 \times 10^{-1} \mathrm{~J}\)
2 \(8 \times 10^{-2} \mathrm{~J}\)
3 \(8 \times 10^{-3} \mathrm{~J}\)
4 \(8 \times 10^{-4} \mathrm{~J}\)
Explanation:
Whenthecapacitor plates get fully charged, there will be no current in branch ab, Remember capacitance acts as the open circuit since capacitance offers infinite resistance to d.c.The capaditance simply collects the charge. Applying Kirchooff's first law to the jundions a and b, we find \(i_{1}=3 A\) and \(i_{2}=1 A\). Now applying Kirchhoff's second law to the closed mesh aefba, weget \(3 \times 5+3 \times 1+1 \times 2=V_{a}-V_{b}\) \(V_{a}-V_{b}=20 \mathrm{~V}\) Energy stored in the capacitor \(U=\frac{1}{2} C\left(V_{a}-V_{b}\right)^{2}=\frac{1}{2} \times 4 \times 10^{-6} \times(20)^{2}\) \(=8 \times 10^{-4} \mathrm{~J}\)
Current Electricity
268536
Equivalent resistance across\(A\) and \(B\) in the given circuit if \(r=10_{\Omega}, R=20 \Omega\) is
