357350
In Wheatstone's network \(P = 2\Omega ,Q = 2\Omega ,R = 2\Omega \,and\,S = 3\Omega \).The resistance with which \(S\) is to shunted in order that the bridge may be balanced is
1 \(2\Omega \)
2 \(6\Omega \)
3 \(1\Omega \)
4 \(4\Omega \)
Explanation:
Let \(X\) be the resistance with which \(S\) is to be shunted for the bridge to be balanced. Then, for balanced Wheatstone's bridge \(\frac{P}{Q} = \frac{R}{{\frac{{SX}}{{S + X}}}}\) (\(\because \) \(S\) and \(X\) are in parallel) Substituting the given values, we get \(\frac{2}{2} = \frac{2}{{\frac{{3X}}{{3 + X}}}}\) or, \(3X = 6 + 2X\) \(\therefore \;\;X = 6\,\Omega \)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
357351
Find the value of \(R\) so that no deflection is noticed in the galvanometer when the switch \(S\) is closed or open
1 \(6\Omega \)
2 \(8\Omega \)
3 \(4\Omega \)
4 None of these
Explanation:
It is a Wheatstone bridge, therefore \(\frac{3}{{3 + 3}} = \frac{{\frac{{1 + 1}}{{8R}}}}{{8 + R}}\,\, \Rightarrow \,R = 8\Omega \)
PHXII03:CURRENT ELECTRICITY
357352
Three resistance \(P\), \(Q\), \(R\) each of \(2\Omega \) and an unknown resistance \(S\) form the four arms of a Wheatstone bridge circuit. When a resistance of \(6\Omega \) is connected in parallel to \(S\) the bridge gets balanced. What is the value of \(S\) ?
1 \(2\Omega \)
2 \(6\Omega \)
3 \(3\Omega \)
4 \(1\Omega \)
Explanation:
As resistance \(S\) and \(6{\kern 1pt} \Omega \) are in parallel their effective resistance is \(\frac{{6S}}{{6 + S}}\,\Omega \) As the bridge is balanced, hence it is balanced Wheatstone bridge. For balancing condition, \(\frac{P}{Q} = \frac{R}{{\left( {\frac{{6S}}{{6 + S}}} \right)}}\quad \Rightarrow \frac{2}{2} = \frac{{2\left( {6 + S} \right)}}{{6S}}\quad \Rightarrow S = 3\Omega \)
PHXII03:CURRENT ELECTRICITY
357353
The value of unknown resistance \((x)\) for which the potential differnce between \(B\) and \(D\) will be zero in the arrangement shown, is
1 \(42\,\Omega \)
2 \(9\,\Omega \)
3 \(6\,\Omega \)
4 \(3\,\Omega \)
Explanation:
The given circuit can be drawn as
Given that, \(V_{B}=V_{D}\), thus the Wheatstone bridge is balanced. \(\therefore \quad \dfrac{P}{Q}=\dfrac{R}{S}\) \({\text{or }}\,\,\,\,\,\frac{{12}}{{6 + x}} = \frac{{0.5}}{{0.5}}\) \( \Rightarrow 6 = 3 + 0.5x \Rightarrow x = 6\,\Omega \)
357350
In Wheatstone's network \(P = 2\Omega ,Q = 2\Omega ,R = 2\Omega \,and\,S = 3\Omega \).The resistance with which \(S\) is to shunted in order that the bridge may be balanced is
1 \(2\Omega \)
2 \(6\Omega \)
3 \(1\Omega \)
4 \(4\Omega \)
Explanation:
Let \(X\) be the resistance with which \(S\) is to be shunted for the bridge to be balanced. Then, for balanced Wheatstone's bridge \(\frac{P}{Q} = \frac{R}{{\frac{{SX}}{{S + X}}}}\) (\(\because \) \(S\) and \(X\) are in parallel) Substituting the given values, we get \(\frac{2}{2} = \frac{2}{{\frac{{3X}}{{3 + X}}}}\) or, \(3X = 6 + 2X\) \(\therefore \;\;X = 6\,\Omega \)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
357351
Find the value of \(R\) so that no deflection is noticed in the galvanometer when the switch \(S\) is closed or open
1 \(6\Omega \)
2 \(8\Omega \)
3 \(4\Omega \)
4 None of these
Explanation:
It is a Wheatstone bridge, therefore \(\frac{3}{{3 + 3}} = \frac{{\frac{{1 + 1}}{{8R}}}}{{8 + R}}\,\, \Rightarrow \,R = 8\Omega \)
PHXII03:CURRENT ELECTRICITY
357352
Three resistance \(P\), \(Q\), \(R\) each of \(2\Omega \) and an unknown resistance \(S\) form the four arms of a Wheatstone bridge circuit. When a resistance of \(6\Omega \) is connected in parallel to \(S\) the bridge gets balanced. What is the value of \(S\) ?
1 \(2\Omega \)
2 \(6\Omega \)
3 \(3\Omega \)
4 \(1\Omega \)
Explanation:
As resistance \(S\) and \(6{\kern 1pt} \Omega \) are in parallel their effective resistance is \(\frac{{6S}}{{6 + S}}\,\Omega \) As the bridge is balanced, hence it is balanced Wheatstone bridge. For balancing condition, \(\frac{P}{Q} = \frac{R}{{\left( {\frac{{6S}}{{6 + S}}} \right)}}\quad \Rightarrow \frac{2}{2} = \frac{{2\left( {6 + S} \right)}}{{6S}}\quad \Rightarrow S = 3\Omega \)
PHXII03:CURRENT ELECTRICITY
357353
The value of unknown resistance \((x)\) for which the potential differnce between \(B\) and \(D\) will be zero in the arrangement shown, is
1 \(42\,\Omega \)
2 \(9\,\Omega \)
3 \(6\,\Omega \)
4 \(3\,\Omega \)
Explanation:
The given circuit can be drawn as
Given that, \(V_{B}=V_{D}\), thus the Wheatstone bridge is balanced. \(\therefore \quad \dfrac{P}{Q}=\dfrac{R}{S}\) \({\text{or }}\,\,\,\,\,\frac{{12}}{{6 + x}} = \frac{{0.5}}{{0.5}}\) \( \Rightarrow 6 = 3 + 0.5x \Rightarrow x = 6\,\Omega \)
357350
In Wheatstone's network \(P = 2\Omega ,Q = 2\Omega ,R = 2\Omega \,and\,S = 3\Omega \).The resistance with which \(S\) is to shunted in order that the bridge may be balanced is
1 \(2\Omega \)
2 \(6\Omega \)
3 \(1\Omega \)
4 \(4\Omega \)
Explanation:
Let \(X\) be the resistance with which \(S\) is to be shunted for the bridge to be balanced. Then, for balanced Wheatstone's bridge \(\frac{P}{Q} = \frac{R}{{\frac{{SX}}{{S + X}}}}\) (\(\because \) \(S\) and \(X\) are in parallel) Substituting the given values, we get \(\frac{2}{2} = \frac{2}{{\frac{{3X}}{{3 + X}}}}\) or, \(3X = 6 + 2X\) \(\therefore \;\;X = 6\,\Omega \)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
357351
Find the value of \(R\) so that no deflection is noticed in the galvanometer when the switch \(S\) is closed or open
1 \(6\Omega \)
2 \(8\Omega \)
3 \(4\Omega \)
4 None of these
Explanation:
It is a Wheatstone bridge, therefore \(\frac{3}{{3 + 3}} = \frac{{\frac{{1 + 1}}{{8R}}}}{{8 + R}}\,\, \Rightarrow \,R = 8\Omega \)
PHXII03:CURRENT ELECTRICITY
357352
Three resistance \(P\), \(Q\), \(R\) each of \(2\Omega \) and an unknown resistance \(S\) form the four arms of a Wheatstone bridge circuit. When a resistance of \(6\Omega \) is connected in parallel to \(S\) the bridge gets balanced. What is the value of \(S\) ?
1 \(2\Omega \)
2 \(6\Omega \)
3 \(3\Omega \)
4 \(1\Omega \)
Explanation:
As resistance \(S\) and \(6{\kern 1pt} \Omega \) are in parallel their effective resistance is \(\frac{{6S}}{{6 + S}}\,\Omega \) As the bridge is balanced, hence it is balanced Wheatstone bridge. For balancing condition, \(\frac{P}{Q} = \frac{R}{{\left( {\frac{{6S}}{{6 + S}}} \right)}}\quad \Rightarrow \frac{2}{2} = \frac{{2\left( {6 + S} \right)}}{{6S}}\quad \Rightarrow S = 3\Omega \)
PHXII03:CURRENT ELECTRICITY
357353
The value of unknown resistance \((x)\) for which the potential differnce between \(B\) and \(D\) will be zero in the arrangement shown, is
1 \(42\,\Omega \)
2 \(9\,\Omega \)
3 \(6\,\Omega \)
4 \(3\,\Omega \)
Explanation:
The given circuit can be drawn as
Given that, \(V_{B}=V_{D}\), thus the Wheatstone bridge is balanced. \(\therefore \quad \dfrac{P}{Q}=\dfrac{R}{S}\) \({\text{or }}\,\,\,\,\,\frac{{12}}{{6 + x}} = \frac{{0.5}}{{0.5}}\) \( \Rightarrow 6 = 3 + 0.5x \Rightarrow x = 6\,\Omega \)
357350
In Wheatstone's network \(P = 2\Omega ,Q = 2\Omega ,R = 2\Omega \,and\,S = 3\Omega \).The resistance with which \(S\) is to shunted in order that the bridge may be balanced is
1 \(2\Omega \)
2 \(6\Omega \)
3 \(1\Omega \)
4 \(4\Omega \)
Explanation:
Let \(X\) be the resistance with which \(S\) is to be shunted for the bridge to be balanced. Then, for balanced Wheatstone's bridge \(\frac{P}{Q} = \frac{R}{{\frac{{SX}}{{S + X}}}}\) (\(\because \) \(S\) and \(X\) are in parallel) Substituting the given values, we get \(\frac{2}{2} = \frac{2}{{\frac{{3X}}{{3 + X}}}}\) or, \(3X = 6 + 2X\) \(\therefore \;\;X = 6\,\Omega \)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
357351
Find the value of \(R\) so that no deflection is noticed in the galvanometer when the switch \(S\) is closed or open
1 \(6\Omega \)
2 \(8\Omega \)
3 \(4\Omega \)
4 None of these
Explanation:
It is a Wheatstone bridge, therefore \(\frac{3}{{3 + 3}} = \frac{{\frac{{1 + 1}}{{8R}}}}{{8 + R}}\,\, \Rightarrow \,R = 8\Omega \)
PHXII03:CURRENT ELECTRICITY
357352
Three resistance \(P\), \(Q\), \(R\) each of \(2\Omega \) and an unknown resistance \(S\) form the four arms of a Wheatstone bridge circuit. When a resistance of \(6\Omega \) is connected in parallel to \(S\) the bridge gets balanced. What is the value of \(S\) ?
1 \(2\Omega \)
2 \(6\Omega \)
3 \(3\Omega \)
4 \(1\Omega \)
Explanation:
As resistance \(S\) and \(6{\kern 1pt} \Omega \) are in parallel their effective resistance is \(\frac{{6S}}{{6 + S}}\,\Omega \) As the bridge is balanced, hence it is balanced Wheatstone bridge. For balancing condition, \(\frac{P}{Q} = \frac{R}{{\left( {\frac{{6S}}{{6 + S}}} \right)}}\quad \Rightarrow \frac{2}{2} = \frac{{2\left( {6 + S} \right)}}{{6S}}\quad \Rightarrow S = 3\Omega \)
PHXII03:CURRENT ELECTRICITY
357353
The value of unknown resistance \((x)\) for which the potential differnce between \(B\) and \(D\) will be zero in the arrangement shown, is
1 \(42\,\Omega \)
2 \(9\,\Omega \)
3 \(6\,\Omega \)
4 \(3\,\Omega \)
Explanation:
The given circuit can be drawn as
Given that, \(V_{B}=V_{D}\), thus the Wheatstone bridge is balanced. \(\therefore \quad \dfrac{P}{Q}=\dfrac{R}{S}\) \({\text{or }}\,\,\,\,\,\frac{{12}}{{6 + x}} = \frac{{0.5}}{{0.5}}\) \( \Rightarrow 6 = 3 + 0.5x \Rightarrow x = 6\,\Omega \)
