A wire of resistance R$R$ is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

Related Formula

Iₛₑᵣᵢₑₛ = (2E)/(R + 2r)$$I_{\text{series}} = \frac{2E}{R + 2r}$$ Iparallel = (E)/(R + r/2)$$I_{\text{parallel}} = \frac{E}{R + r/2}$$

Core Logic

Since the external current is identical in both cases, equate the series current expression to the parallel current expression.

Step 1: Setting up equation

(2E)/(6 + 2r) = (E)/(6 + (r)/(2))$$\frac{2E}{6 + 2r} = \frac{E}{6 + \frac{r}{2}}$$

Step 2: Solving for r

Divide both sides by E$E$ and cross-multiply:

2 (6 + (r)/(2)) = 6 + 2r$$2 \left(6 + \frac{r}{2}\right) = 6 + 2r$$

12 + r = 6 + 2r$12 + r = 6 + 2r$ 12 - 6 = 2r - r$12 - 6 = 2r - r$ r = 6 Ω$r = 6 \,\Omega$

Pattern Recognition

A well-known standard shortcut: If n$n$ identical cells yield the same current in series and parallel across an external resistor R$R$, then r = R$r = R$. Here R = 6$R = 6$, so immediately r = 6$r = 6$.

Chapter Mix

Class 12 Physics: Current Electricity

Related Formula

Kirchhoff's Voltage Law (KVL): Σ V = 0$$\text{Kirchhoff's Voltage Law (KVL): } \sum V = 0$$ V = I · Req$$V = I \cdot R_{\text{eq}}$$

Core Logic

This is an unbalanced Wheatstone bridge. We can solve it by connecting an imaginary 1V battery across AB, applying Kirchhoff's laws to find the total current I$I$, and then finding Req = V / I$R_{\text{eq}} = V / I$.

Step 1: First KVL Loop

Let total current from the 1V source be i$i$. It splits into i₁$i_1$ through the upper 6 Ω$6\,\Omega$ resistor, and (i - i₁)$(i - i_1)$ through the lower 3 Ω$3\,\Omega$ resistor. Applying KVL to the upper closed loop of the bridge:

6i₁ + 3(2i₁ - i) = 3(i - i₁)$$6i_1 + 3(2i_1 - i) = 3(i - i_1)$$ 6i₁ + 6i₁ - 3i = 3i - 3i₁$$6i_1 + 6i_1 - 3i = 3i - 3i_1$$ 15i₁ = 6i ⇒ i₁ = (2)/(5)i --- (1)$$15i_1 = 6i \Rightarrow i_1 = \frac{2}{5}i \quad \text{--- (1)}$$

Step 2: Second KVL Loop (Outer Battery Loop)

Apply KVL through the lower path and the battery:

3(i - i₁) + 6(current in next branch) = 1$$3(i - i_1) + 6(\text{current in next branch}) = 1$$

Wait, looking at the node splits in the solution: the current entering B from the bottom resistor is i₁$i_1$, and from top is (i - i₁)$(i - i_1)$ due to symmetry. Equation from the lower path:

3(i - i₁) + 6i₁ = 1$$3(i - i_1) + 6i_1 = 1$$

Substitute i₁ = (2)/(5)i$i_1 = \frac{2}{5}i$: 3i + 3i₁ = 1$3i + 3i_1 = 1$

3i + 3((2)/(5)i) = 1$$3i + 3\left(\frac{2}{5}i\right) = 1$$ i (3 + (6)/(5)) = 1$$i \left(3 + \frac{6}{5}\right) = 1$$ i ((21)/(5)) = 1 ⇒ i = (5)/(21) ~A$$i \left(\frac{21}{5}\right) = 1 \Rightarrow i = \frac{5}{21} \mathrm{~A}$$

Step 3: Finding Equivalent Resistance

Req = (V)/(i) = (1)/(5 / 21) = (21)/(5) Ω$$R_{\text{eq}} = \frac{V}{i} = \frac{1}{5 / 21} = \frac{21}{5} \,\Omega$$

Given Req = (x)/(5) Ω$R_{\text{eq}} = \frac{x}{5} \,\Omega$. Therefore, x = 21$x = 21$.

Pattern Recognition

For a symmetrical unbalanced bridge (like R₁=6, R₂=3$R_1=6, R_2=3$ and R₃=3, R₄=6$R_3=3, R_4=6$), the currents divide symmetrically. Assuming a 1V source and tracking current i$i$ and i₁$i_1$ is the fastest fail-proof algebraic method.

Chapter Mix

Class 12 Physics: Current Electricity

Core Logic

Initially, R₁ = R₂ = R₃ = R₄ = R$R_{1} = R_{2} = R_{3} = R_{4} = R$. The bridge is balanced, and VA = VB$V_A = V_B$. After heating, R₃$R_3$ increases by 10%, meaning the new R₃' = R + 0.1R = 1.1R$R_3' = R + 0.1R = 1.1R$. The other arms remain R$R$. The battery provides a voltage of V = 40 V$V = 40\text{ V}$.

Step 1: Calculate Potentials at nodes A and B

Let the lower node (D) be at 0 V$0\text{ V}$ and the upper node (C) be at 40 V$40\text{ V}$. The potential at node A (between R₃'$R_3'$ and R₄$R_4$) can be found via the voltage divider rule across the right branch (R₃'$R_3'$ and R₄$R_4$).

Wait, examining the circuit labels carefully: Assume the branches are: Left branch contains R₁$R_1$ and R₂$R_2$. Right branch contains R₃$R_3$ and R₄$R_4$. Node B is between R₁$R_1$ and R₂$R_2$. Node A is between R₃$R_3$ and R₄$R_4$.

From the provided solution:

VA = (V)/(2)$$V_A = \frac{V}{2}$$

This implies the branch containing node A consists of two equal resistances R$R$, which contradicts the textual convention if R₃$R_3$ changed. Looking at the provided solution image logic, the branch with B$B$ has the changed resistance 1.1R$1.1R$. Let's follow the PDF's mathematical layout exactly.

VA = (V)/(2)$$V_A = \frac{V}{2}$$ (Meaning branch A has equal resistors R$R$ and R$R$)

VB = (V)/(R + 1.1R) × R = (V)/(2.1)$$V_B = \frac{V}{R + 1.1R} \times R = \frac{V}{2.1}$$

(Note: The PDF solution computes VB = V / 2.1$V_B = V / 2.1$, implying the resistance below node B is R$R$, and above is 1.1R$1.1R$)

Step 2: Potential Difference Calculation

VA - VB = V [ (1)/(2) - (1)/(2.1) ]$$\therefore V_A - V_B = V \left[ \frac{1}{2} - \frac{1}{2.1} \right]$$ VA - VB = 40 × [ (2.1 - 2)/(2 × 2.1) ]$$V_A - V_B = 40 \times \left[ \frac{2.1 - 2}{2 \times 2.1} \right]$$ VA - VB = (0.1)/(4.2) × 40 = (4)/(4.2)$$V_A - V_B = \frac{0.1}{4.2} \times 40 = \frac{4}{4.2}$$ VA - VB = (40)/(42) ≈ 0.952 V$$V_A - V_B = \frac{40}{42} \approx 0.952 \text{ V}$$

Step 3: Final Answer

The closest match among the options is 0.95 V$0.95\text{ V}$.

Pattern Recognition

For a slightly unbalanced Wheatstone bridge due to a fractional change Δ R/R = x$\Delta R/R = x$, the potential difference Δ V$\Delta V$ is approximately V · (x)/(4)$V \cdot \frac{x}{4}$ for small x$x$. Here x=0.1$x=0.1$, so Δ V ≈ 40 × 0.1 / 4 = 1 V$\Delta V \approx 40 \times 0.1 / 4 = 1 \text{ V}$. Exact calculation gives 0.95 V$0.95\text{ V}$.

Chapter Mix

Class 12 Physics: Current Electricity

Related Formula

E = V · Q$E = V \cdot Q$

Q = I · t$Q = I \cdot t$

Core Logic

The capacity rating 5800~mAh$5800\mathrm{~mAh}$ represents total electric charge Q$Q$ stored:

Q = 5800~mA × 1~hour = (5800 × 10⁻³~A) × 3600~s = 20880~C$$Q = 5800\mathrm{~mA} \times 1\mathrm{~hour} = (5800 \times 10^{-3}\mathrm{~A}) \times 3600\mathrm{~s} = 20880\mathrm{~C}$$

Given the voltage is V = 4.2~V$V = 4.2\mathrm{~V}$, the energy stored in Joules is:

E = V · Q = 4.2 × 20880 = 87696~J = 87.696~kJ ≈ 87.7~kJ$$E = V \cdot Q = 4.2 \times 20880 = 87696\mathrm{~J} = 87.696\mathrm{~kJ} \approx 87.7\mathrm{~kJ}$$

Step 1: Final Conclusion

The electrical energy stored in the fully charged battery is 87.7~kJ$87.7\mathrm{~kJ}$.

Pattern Recognition

Always remember: Energy (in Joules) = Voltage (V) × Capacity (Ah) × 3600$\text{Energy (in Joules)} = \text{Voltage (V)} \times \text{Capacity (Ah)} \times 3600$. This directly transforms electrical rating units to standard SI energy units.

Chapter Mix

Class 12 Physics: Current Electricity

Related Formula

• Series cell configuration:

εeq = ε₁ + ε₂, req = r₁ + r₂$$\varepsilon_{\text{eq}} = \varepsilon_1 + \varepsilon_2, \quad r_{\text{eq}} = r_1 + r_2$$

• Parallel cell configuration (for unequal cells in parallel):

εeq = ((ε₁)/(r₁) + (ε₂)/(r₂))/((1)/(r₁) + (1)/(r₂)), 1req = (1)/(r₁) + (1)/(r₂)$$\varepsilon_{\text{eq}} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}, \quad \frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}$$

• Circuit current:

I = εeqreq + R$$I = \frac{\varepsilon_{\text{eq}}}{r_{\text{eq}} + R}$$

Core Logic

Given parameters:

• Cell 1:$

  Core Logic

  Given parameters:

  • Cell 1: $\varepsilon_1 = 1\mathrm{~V},$, $r_1 = 2\Omega
  • Cell 2:$
  • Cell 2: $\varepsilon_2 = 2\mathrm{~V},$, $r_2 = 1\Omega
  • External resistance$
  • External resistance $R = 6\Omega

  Step 1: Calculate$

Step 1: Calculate $I_1(Series Configuration)

$ (Series Configuration)

$εeq = 1 + 2 = 3~V$\varepsilon_{\text{eq}} = 1 + 2 = 3\mathrm{~V}$$ $req = 2 + 1 = 3Ω$r_{\text{eq}} = 2 + 1 = 3\Omega$$ $I₁ = (3)/(3 + 6) = (3)/(9) = (1)/(3)~A$I_1 = \frac{3}{3 + 6} = \frac{3}{9} = \frac{1}{3}\mathrm{~A}$

Step 2: Calculate $I₂$ (Parallel Configuration)

$$

Step 2: Calculate $I_2$ (Parallel Configuration)

$$\varepsilon_{\text{eq}} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{0.5 + 2}{1.5} = \frac{2.5}{1.5} = \frac{5}{3}\mathrm{~V}$$ $$r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\Omega$$ $$I_2 = \frac{\varepsilon_{\text{eq}}}{r_{\text{eq}} + R} = \frac{\frac{5}{3}}{\frac{2}{3} + 6} = \frac{\frac{5}{3}}{\frac{20}{3}} = \frac{5}{20} = \frac{1}{4}\mathrm{~A}$

Step 3: Find ratio and evaluate $x$

$$$

Step 3: Find ratio and evaluate $x$

$$\frac{I_1}{I_2} = \frac{\frac{1}{3}}{\frac{1}{4}} = \frac{4}{3}$

Comparing with$

Comparing with $\frac{x}{3}:

$:

$x = 4

Pattern Recognition

For cells connected in parallel, the formula for equivalent EMF$

Pattern Recognition

For cells connected in parallel, the formula for equivalent EMF $\varepsilon_{\text{eq}}$ can be viewed as a weighted average. Connecting cells in series maximizes EMF but increases internal resistance, while parallel configuration limits EMF to an intermediate value while reducing equivalent internal resistance.

Chapter Mix

Class 12 Physics: Current Electricity