1d:I[72985,["/_next/static/chunks/3th5o5azza5j1.js","/_next/static/chunks/0953gtd9gbswy.js","/_next/static/chunks/01oly6y_17ki0.js","/_next/static/chunks/1xlwfijmtjgqj.js","/_next/static/chunks/2hzca_9a7jrui.js","/_next/static/chunks/0_v6xzhg0htsf.js","/_next/static/chunks/2x0yuymt3nfdv.js","/_next/static/chunks/1hjnhv8wc11is.js"],"default"] 1e:I[63508,["/_next/static/chunks/3th5o5azza5j1.js","/_next/static/chunks/0953gtd9gbswy.js","/_next/static/chunks/01oly6y_17ki0.js","/_next/static/chunks/1xlwfijmtjgqj.js","/_next/static/chunks/2hzca_9a7jrui.js","/_next/static/chunks/0_v6xzhg0htsf.js","/_next/static/chunks/2x0yuymt3nfdv.js","/_next/static/chunks/1hjnhv8wc11is.js"],"default"] 1f:T1313,

An electrical circuit consists of two resistors: the first resistor ( $R_1$ ) has a resistance of 10 Ω and the second resistor ( $R_2$ ) has a resistance of 40 Ω. Calculate the total equivalent resistance of the circuit when the two resistors are connected in series ( $R_{eq,s}$ ) and when they are connected in parallel ( $R_{eq,p}$ ). (Note: You can modify the variables in the input data section)

20:Tf3c,

Click to view Series vs. Parallel Circuit comparison table

Key electrical differences for quick analysis and answer verification:

Electrical Property	Series Circuit	Parallel Circuit
Equivalent resistance (2 resistors)	R_eq,s = R₁ + R₂	R_eq,p = (R₁ · R₂) / (R₁ + R₂)
Magnitude check rule	Always greater than the largest resistor	Always smaller than the smallest resistor
Electric current (I)	Same through all components (I = I₁ = I₂)	Divides among branches (I = I₁ + I₂)
Voltage drop (V)	Divides proportionally (V = V₁ + V₂)	Same across all branches (V = V₁ = V₂)

21:T49dd,

Step 1: Calculate the equivalent resistance in series ( $R_{eq,s}$ ) (Directly sum the individual resistances since the same current flows sequentially through a single path)

Step 2: Calculate the equivalent resistance in parallel ( $R_{eq,p}$ ) (Apply the product-over-sum formula for two resistors connected across the same pair of nodes)

💡 Calculator Tip: When calculating parallel resistance using the reciprocal formula $\frac{1}{R_{eq,p}} = \frac{1}{R_1} + \frac{1}{R_2}$ , entering $\frac{1}{10} + \frac{1}{40}$ on a calculator gives $0.125\ \Omega^{-1}$ , which is the conductance rather than the resistance. You must invert the sum in the final step ( $R_{eq,p} = \frac{1}{0.125} = 8$ Ω), or use the product-over-sum formula $(10 \times 40) \div (10 + 40) = \frac{400}{50} = 8$ Ω directly to avoid forgetting the final inversion.

22:Tec4,

From the first step calculation, connecting the two resistors in series yields an equivalent resistance ( $R_{eq,s}$ ) of , which is always greater than the largest individual resistor in the chain (40 Ω). Because electric current must flow sequentially through a single path, adding resistors in series effectively lengthens the resistive path. In engineering practice, series configurations are widely used to limit circuit current or create voltage divider networks, with the trade-off that if any single component fails open, current stops flowing through the entire loop.

Conversely, in the second step, connecting the exact same pair of resistors in parallel reduces the equivalent resistance ( $R_{eq,p}$ ) to just , which is always smaller than the smallest individual branch resistor (10 Ω). Providing multiple parallel branches makes it easier for total current to flow through the network. For this reason, electrical engineers wire household and building outlets and lighting loads in parallel so every appliance receives the same standard supply voltage (e.g., 220 V) and continues operating normally even if another device is switched off or fails.

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This example demonstrates how to calculate the equivalent resistance of an electrical circuit when resistors are connected in series and in parallel, simplifying multiple resistors into a single total resistance for straightforward circuit analysis.

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