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To determine the dimensional formulae of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] given the expression for the electric current [tex]\( I \)[/tex] in a circuit, [tex]\( I = \frac{V}{A}\left(1 - e^{-Bt}\right) \)[/tex], we need to analyze the dimensions for each term.
1. Dimension of Current (I): The electric current [tex]\( I \)[/tex] is measured in amperes (A). Thus, its dimensional formula is:
[tex]\[ [I] = A \][/tex]
2. Dimension of Voltage (V): The voltage [tex]\( V \)[/tex] has the dimensional formula of:
[tex]\[ [V] = M L^2 T^{-3} A^{-1} \][/tex]
This is derived from the formula [tex]\( V = W/Q \)[/tex], where [tex]\( W \)[/tex] is work (ML^2 T^{-2}) and [tex]\( Q \)[/tex] is charge (A T).
3. Expression Analysis: For the equation [tex]\( I = \frac{V}{A} \left(1 - e^{-Bt}\right) \)[/tex], the term inside the parentheses, [tex]\( \left(1 - e^{-Bt}\right) \)[/tex], is dimensionless because the exponential function and the constant 1 must be dimensionless to be subtracted.
Therefore, to keep the equation dimensionally consistent, the term [tex]\( \frac{V}{A} \)[/tex] must have the same dimensions as current [tex]\( I \)[/tex] (A).
[tex]\[ \frac{V}{A} \text{ must have the same dimensions as } [I] = A \][/tex]
Given [tex]\( [V] = M L^2 T^{-3} A^{-1} \)[/tex]:
[tex]\[ \frac{[V]}{[A]} = \frac{M L^2 T^{-3} A^{-1}}{[A]} \][/tex]
where [tex]\( [A] \)[/tex] here represents the dimensional formula we are seeking. To maintain dimensional consistency:
[tex]\[ \frac{M L^2 T^{-3} A^{-1}}{[A]} = A \][/tex]
4. Solve for [tex]\( [A] \)[/tex]:
Let’s denote the dimensional formula of [tex]\( A \)[/tex] symbolically as [tex]\( [A] = M^a L^b T^c A^d \)[/tex].
Equating the dimensions, we get:
[tex]\[ \frac{M L^2 T^{-3} A^{-1}}{M^a L^b T^c A^d} = A \][/tex]
This simplifies to:
[tex]\[ M^{1-a} L^{2-b} T^{-3-c} A^{-1-d} = A \][/tex]
Thus, equating each dimension:
[tex]\[ M^{1-a} = 1 \implies 1 - a = 0 \implies a = 1 \][/tex]
[tex]\[ L^{2-b} = 1 \implies 2 - b = 0 \implies b = 2 \][/tex]
[tex]\[ T^{-3-c} = 1 \implies -3 - c = 0 \implies c = -3 \][/tex]
[tex]\[ A^{-1-d} = A \implies -1 - d = 1 \implies d = -2 \][/tex]
Therefore, the dimensional formula of [tex]\( A \)[/tex] is:
[tex]\[ [A] = M^1 L^2 T^{-3} A^{-2} = M L^2 T^{-3} A^{-2} \][/tex]
5. Dimension of B: Since the term [tex]\( B t \)[/tex] inside the exponential must be dimensionless, [tex]\( B \)[/tex] must have dimensions inverse to that of time [tex]\( t \)[/tex].
[tex]\[ [B t] = 1 \implies [B] \cdot [t] = 1 \][/tex]
Given that the dimension of time [tex]\( [t] \)[/tex] is [tex]\( T \)[/tex]:
[tex]\[ [B] \cdot T = 1 \implies [B] = \frac{1}{T} = T^{-1} \][/tex]
Summarizing:
The dimensional formula of [tex]\( A \)[/tex] is [tex]\( M L^2 T^{-3} A^{-2} \)[/tex] and the dimensional formula of [tex]\( B \)[/tex] is [tex]\( T^{-1} \)[/tex].
Thus, the correct answer is:
(3) [tex]\(\left[M L^2 T^{-3} A^{-2}\right],\left[T^{-1}\right]\)[/tex]
1. Dimension of Current (I): The electric current [tex]\( I \)[/tex] is measured in amperes (A). Thus, its dimensional formula is:
[tex]\[ [I] = A \][/tex]
2. Dimension of Voltage (V): The voltage [tex]\( V \)[/tex] has the dimensional formula of:
[tex]\[ [V] = M L^2 T^{-3} A^{-1} \][/tex]
This is derived from the formula [tex]\( V = W/Q \)[/tex], where [tex]\( W \)[/tex] is work (ML^2 T^{-2}) and [tex]\( Q \)[/tex] is charge (A T).
3. Expression Analysis: For the equation [tex]\( I = \frac{V}{A} \left(1 - e^{-Bt}\right) \)[/tex], the term inside the parentheses, [tex]\( \left(1 - e^{-Bt}\right) \)[/tex], is dimensionless because the exponential function and the constant 1 must be dimensionless to be subtracted.
Therefore, to keep the equation dimensionally consistent, the term [tex]\( \frac{V}{A} \)[/tex] must have the same dimensions as current [tex]\( I \)[/tex] (A).
[tex]\[ \frac{V}{A} \text{ must have the same dimensions as } [I] = A \][/tex]
Given [tex]\( [V] = M L^2 T^{-3} A^{-1} \)[/tex]:
[tex]\[ \frac{[V]}{[A]} = \frac{M L^2 T^{-3} A^{-1}}{[A]} \][/tex]
where [tex]\( [A] \)[/tex] here represents the dimensional formula we are seeking. To maintain dimensional consistency:
[tex]\[ \frac{M L^2 T^{-3} A^{-1}}{[A]} = A \][/tex]
4. Solve for [tex]\( [A] \)[/tex]:
Let’s denote the dimensional formula of [tex]\( A \)[/tex] symbolically as [tex]\( [A] = M^a L^b T^c A^d \)[/tex].
Equating the dimensions, we get:
[tex]\[ \frac{M L^2 T^{-3} A^{-1}}{M^a L^b T^c A^d} = A \][/tex]
This simplifies to:
[tex]\[ M^{1-a} L^{2-b} T^{-3-c} A^{-1-d} = A \][/tex]
Thus, equating each dimension:
[tex]\[ M^{1-a} = 1 \implies 1 - a = 0 \implies a = 1 \][/tex]
[tex]\[ L^{2-b} = 1 \implies 2 - b = 0 \implies b = 2 \][/tex]
[tex]\[ T^{-3-c} = 1 \implies -3 - c = 0 \implies c = -3 \][/tex]
[tex]\[ A^{-1-d} = A \implies -1 - d = 1 \implies d = -2 \][/tex]
Therefore, the dimensional formula of [tex]\( A \)[/tex] is:
[tex]\[ [A] = M^1 L^2 T^{-3} A^{-2} = M L^2 T^{-3} A^{-2} \][/tex]
5. Dimension of B: Since the term [tex]\( B t \)[/tex] inside the exponential must be dimensionless, [tex]\( B \)[/tex] must have dimensions inverse to that of time [tex]\( t \)[/tex].
[tex]\[ [B t] = 1 \implies [B] \cdot [t] = 1 \][/tex]
Given that the dimension of time [tex]\( [t] \)[/tex] is [tex]\( T \)[/tex]:
[tex]\[ [B] \cdot T = 1 \implies [B] = \frac{1}{T} = T^{-1} \][/tex]
Summarizing:
The dimensional formula of [tex]\( A \)[/tex] is [tex]\( M L^2 T^{-3} A^{-2} \)[/tex] and the dimensional formula of [tex]\( B \)[/tex] is [tex]\( T^{-1} \)[/tex].
Thus, the correct answer is:
(3) [tex]\(\left[M L^2 T^{-3} A^{-2}\right],\left[T^{-1}\right]\)[/tex]
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