Question:
In the given figure:
\[
V_1=V,\quad V_2=\alpha V,\quad R_1=\beta R,\quad R_2=\gamma R,
\]
where \(\alpha,\beta,\gamma\) are positive real numbers. The value of current \(I\) is
In the given figure:
\[
V_1=V,\quad V_2=\alpha V,\quad R_1=\beta R,\quad R_2=\gamma R,
\]
where \(\alpha,\beta,\gamma\) are positive real numbers. The value of current \(I\) is
Show Hint
In bridge-type resistor circuits, first identify symmetry and then apply Kirchhoff's Voltage Law carefully. Express all resistances and voltages in terms of the given parameters before simplifying.
Updated On: Jun 25, 2026
- \(\dfrac{(\alpha-1)\gamma}{4\beta(\beta+\gamma)}\dfrac{V}{R}\)
- \(\dfrac{(\alpha-1)}{4\beta}\dfrac{V}{R}\)
- \(\dfrac{(\alpha-1)\beta}{2\gamma(\beta+\gamma)}\dfrac{V}{R}\)
- \(\dfrac{(\alpha-1)(\beta+\gamma)}{2\beta\gamma}\dfrac{V}{R}\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Understand the symmetry of the circuit.
The circuit contains four equal resistors of value \[ R_1=\beta R \] and a central branch containing \[ R_2=\gamma R \] along with a source \(V_2=\alpha V\).
Using Kirchhoff's laws and symmetry, let the current in the right lower branch be \(I\).
Step 2: Apply Kirchhoff's Voltage Law.
The equivalent analysis of the bridge gives the current through the indicated branch as \[ I= \frac{V_2-V_1}{4R_1\left(1+\dfrac{R_1}{R_2}\right)} \] Substituting \[ V_1=V,\qquad V_2=\alpha V, \] we get \[ I= \frac{\alpha V-V}{4\beta R\left(1+\dfrac{\beta R}{\gamma R}\right)} \] \[ = \frac{(\alpha-1)V}{4\beta R\left(1+\dfrac{\beta}{\gamma}\right)} \]
Step 3: Simplify the denominator.
\[ 1+\frac{\beta}{\gamma} = \frac{\beta+\gamma}{\gamma} \] Hence, \[ I= \frac{(\alpha-1)V}{4\beta R\left(\dfrac{\beta+\gamma}{\gamma}\right)} \] \[ = \frac{(\alpha-1)\gamma}{4\beta(\beta+\gamma)} \cdot \frac{V}{R} \]
Step 4: Final conclusion.
Therefore, the current is \[ \boxed{ \dfrac{(\alpha-1)\gamma}{4\beta(\beta+\gamma)} \dfrac{V}{R} } \]
The circuit contains four equal resistors of value \[ R_1=\beta R \] and a central branch containing \[ R_2=\gamma R \] along with a source \(V_2=\alpha V\).
Using Kirchhoff's laws and symmetry, let the current in the right lower branch be \(I\).
Step 2: Apply Kirchhoff's Voltage Law.
The equivalent analysis of the bridge gives the current through the indicated branch as \[ I= \frac{V_2-V_1}{4R_1\left(1+\dfrac{R_1}{R_2}\right)} \] Substituting \[ V_1=V,\qquad V_2=\alpha V, \] we get \[ I= \frac{\alpha V-V}{4\beta R\left(1+\dfrac{\beta R}{\gamma R}\right)} \] \[ = \frac{(\alpha-1)V}{4\beta R\left(1+\dfrac{\beta}{\gamma}\right)} \]
Step 3: Simplify the denominator.
\[ 1+\frac{\beta}{\gamma} = \frac{\beta+\gamma}{\gamma} \] Hence, \[ I= \frac{(\alpha-1)V}{4\beta R\left(\dfrac{\beta+\gamma}{\gamma}\right)} \] \[ = \frac{(\alpha-1)\gamma}{4\beta(\beta+\gamma)} \cdot \frac{V}{R} \]
Step 4: Final conclusion.
Therefore, the current is \[ \boxed{ \dfrac{(\alpha-1)\gamma}{4\beta(\beta+\gamma)} \dfrac{V}{R} } \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Kirchhoff's Laws Questions
In the given circuit, the reading of the ammeter is:
- In a Wheatstone bridge, the bridge is balanced when the ratio of the resistances in the four arms satisfies [H] [width=0.5\linewidth]{108.png}
In the figure, if $V_A = 9V, V_D = 5V, V_C = 7V$, then the current through the wire BD is
- The current \(i\) in the circuit shown in the figure is
- In the given circuit values of \(I_1\), \(I_2\), \(I_3\) are respectively
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions