Let l, r, c, and v represent inductance, resistance, capacitance and voltage, respectively. The dime

Q: Why can't I just substitute the dimensions of $l, r, c,$ and $v$ directly?

You absolutely can! However, finding and substituting the individual dimensions for inductance, resistance, capacitance, and voltage is highly prone to calculation errors and takes a lot of time. Grouping terms is a much faster and more elegant method.

Q: How do I know which terms to group together?

Look for familiar combinations from standard physics formulas. Whenever you see $L$ and $R$ together, think of the time constant $\tau = L/R$. When you see $C$ and $V$, think of the charge $Q = CV$. Practice will make spotting these patterns second nature.

Q: What is the dimension of the time constant $l/r$?

The time constant of an LR circuit represents the time it takes for the current to reach a certain percentage of its maximum value. Therefore, its dimension is simply time, $[T]$.

Q: Why is the dimension of charge $[A \cdot T]$?

Electric current is defined as the rate of flow of charge, $I = Q/t$. Rearranging this gives $Q = I \cdot t$. Since the fundamental dimension for current is Ampere $[A]$ and for time is $[T]$, the dimension of charge is $[A \cdot T]$.

Q: Does this grouping trick work for all dimensional analysis problems?

Not all, but many! JEE often designs questions specifically to reward students who can spot these familiar groupings. Always take a few seconds to look for patterns before diving into brute-force calculations.

Question

Let $l, r, c$, and $v$ represent inductance, resistance, capacitance and voltage, respectively. The dimension of $\frac{l}{rcv}$ in SI units will be

Options:

(A) $[LT^2]$

(B) $[LTA]$

(C) $[A^{-1}]$

(D) $[LA^{-2}]$

Sigma Percentile · Accepted Answer

### The Brute Force Trap

Imagine you are sitting in the exam hall, and you see this expression: $\frac{l}{rcv}$. Your first instinct might be to recall or derive the dimensional formula for each individual quantity.

You would write down the dimension of inductance $l$, then resistance $r$, then capacitance $c$, and finally voltage $v$.

While this method is perfectly valid, it is a massive trap! It is highly prone to algebraic errors and will consume precious minutes that you could spend on harder problems.

### The Smart Approach: Grouping Terms

There is a much more elegant way to solve this. Instead of looking at four isolated variables, we can group them into familiar physical quantities.

JEE problem setters love to design questions that reward students who can spot these hidden patterns. Let's break down the expression $\frac{l}{r \cdot c \cdot v}$ by grouping the terms strategically.

### Analyzing the LR Time Constant

Look closely at the numerator and the first term in the denominator. We have $l$ divided by $r$.

If you recall your study of electromagnetic induction, the ratio $\frac{l}{r}$ is exactly the **time constant** ($	au$) of an LR circuit.

The time constant simply represents a duration of time. Therefore, its dimensional formula is incredibly simple:

$ \left[ \frac{l}{r} ight] = [T] $

### Uncovering the Hidden Charge

Now, let's look at the remaining terms in the denominator: $c$ and $v$.

From electrostatics, we know that the product of capacitance and voltage gives the total **charge** stored in a capacitor. The formula is $q = c \cdot v$.

Since electric current is the rate of flow of charge ($I = \frac{q}{t}$), we can express charge as current multiplied by time. Thus, its dimension is:

$ [c \cdot v] = [q] = [A \cdot T] $

### The Final Assembly

Now, let's bring it all together. We substitute our grouped dimensions back into the original expression.

The complex fraction simplifies beautifully into the ratio of our two new dimensions:

$ \left[ \frac{l}{rcv} ight] = \frac{[T]}{[A \cdot T]} $

Notice how the time dimension $[T]$ appears in both the numerator and the denominator. They perfectly cancel each other out!

We are left with just the inverse of the current dimension.

$ \left[ \frac{l}{rcv} ight] = [A^{-1}] $

And just like that, without writing a single complex dimensional formula, we have arrived at the correct answer. Always keep an eye out for these elegant shortcuts!

Animated Solution for Physics - Physics and Measurement: Let $l, r, c$ , and $v$ represent inductance, resistance, capacitance and voltage, respectively. The dimension of $\frac{l}{r c v}$ in SI units will be

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Visualized Solution

$The Expression$

$The Smart Approach$

$LR Time Constant$

$Charge from Capacitance$

$Substitution$

$Final Answer$

$The Way Forward$

The Sigma Insight: Dimensional Analysis

The Brute Force Trap

The Smart Approach

Analyzing the LR Time Constant

Uncovering the Hidden Charge

The Final Assembly

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Smart Approach

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LR Time Constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Charge from Capacitance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward