the area of triangle opq with o(0 0) p(2 0) and q(0 2) is
Answers
Answer:
area of triangle = 1/2 × b × h
Answer:
Given :-
C is the capacitance of a capacitor discharging through a resistor R.
t₁ is the time taken for the energy stored in the capacitor to reduce to half its initial value
t₂ is the time taken for the charge to reduce to one-fourth its initial value.
⇝ To Find :-
Value of t₁ / t₂
⇝ Solution :-
❒ We Know equation of discharging of a charge is :
\bf q = q_{ \circ} {e}^{-t/RC} \: - - - (1)q=q
∘
e
−t/RC
−−−(1)
★ According To Question :
\begin{gathered} \pink {\text{At time t = t₂ }} \\ \end{gathered}
At time t = t₂
\begin{gathered}\text q = \dfrac{\text q_{ \circ}}{4} \\ \end{gathered}
q=
4
q
∘
✏ Using (1) :
\begin{gathered}:\longmapsto\dfrac {\cancel{\text q_{ \circ}}}{4} = \cancel\text q_{ \circ} {\text e}^{ - { \text t_2/ \text{RC}}} \\ \end{gathered}
:⟼
4
q
∘
=
q
∘
e
−t
2
/RC
\begin{gathered} :\longmapsto\sf\dfrac{1}{4} = {e}^{-t_2/RC} \\ \end{gathered}
:⟼
4
1
=e
−t
2
/RC
★ Taking Log Both Side :
\begin{gathered}:\longmapsto \sf \log \bigg( \dfrac{1}{4} \bigg) = \log \big( {e}^{-t_2/RC} \big) \\ \end{gathered}
:⟼log(
4
1
)=log(e
−t
2
/RC
)
\begin{gathered}:\longmapsto \sf \log {(4)}^{ - 1} = \log( {e)}^{-t_2/RC} \\ \end{gathered}
:⟼log(4)
−1
=log(e)
−t
2
/RC
\begin{gathered}:\longmapsto \sf \cancel- 1( \log4) =\cancel - \dfrac{t_2}{RC}. \log \: e \\ \end{gathered}
:⟼
−
1(log4)=
−
RC
t
2
.loge
\begin{gathered}:\longmapsto \sf \log4 = \dfrac{t_2}{RC} \\ \end{gathered}
:⟼log4=
RC
t
2
\begin{gathered}:\longmapsto \sf t_2 = RC .\log4 \\ \end{gathered}
:⟼t
2
=RC.log4
\begin{gathered}:\longmapsto \sf t_2 = RC . \log{2}^{2} \\ \end{gathered}
:⟼t
2
=RC.log2
2
:\longmapsto \boxed{ \pink{ \boxed{\bf t_2 = 2\times RC \times log2}}}:⟼
t
2
=2×RC×log2
❒As Energy Stored in a capacitor is :
\begin{gathered} \text E = \frac{1}{2} \dfrac{ {\text q_{ \circ}}^{2} }{ \text{C}} \: \: - - - (2) \\ \end{gathered}
E=
2
1
C
q
∘
2
−−−(2)
★ Let when charge is q₁ , Energy becomes E/2
So,
\begin{gathered} \sf \dfrac{E}{2} = \frac{1}{2} \frac{q_1 {}^{2} }{C} \\ \end{gathered}
2
E
=
2
1
C
q
1
2
✏ Using (2) :
\begin{gathered}:\longmapsto \sf \cancel\dfrac{1}{2} \frac{ {q_{ \circ}}^{2} }{2 \cancel C } =\cancel\dfrac{1}{2} \frac{ {q_1}^{2} }{C} \\ \end{gathered}
:⟼
2
1
2
C
q
∘
2
=
2
1
C
q
1
2
:\longmapsto \bf q_1 = \dfrac{q_{ \circ}}{\sqrt{2}}\:\: ----(3):⟼q
1
=
2
q
∘
−−−−(3)
✏ Using (1) :
\begin{gathered}:\longmapsto \sf q_1 = q_{ \circ} { \: e}^{ - t_1/RC} \\ \end{gathered}
:⟼q
1
=q
∘
e
−t
1
/RC
✏ Using (3) :
\begin{gathered}:\longmapsto \sf \dfrac{ \cancel {q_{ \circ}}}{ \sqrt{2} } = \cancel{q_{ \circ}} \: {e}^{ - t_1/RC} \\ \end{gathered}
:⟼
2
q
∘
=
q
∘
e
−t
1
/RC
\begin{gathered}:\longmapsto \sf \dfrac{1}{ \sqrt{2} } = {e}^{ - t_1/RC} \\ \end{gathered}
:⟼
2
1
=e
−t
1
/RC
★ Taking Log Both Side :
\begin{gathered}:\longmapsto \sf\log \bigg( \dfrac{1}{ \sqrt{2} } \bigg) = \log( {e)}^{ - t_1/RC} \\ \end{gathered}
:⟼log(
2
1
)=log(e)
−t
1
/RC
\begin{gathered}:\longmapsto \sf\log(2) {}^{ - 1/2} = \log( {e)}^{ - t_1/RC} \\ \end{gathered}
:⟼log(2)
−1/2
=log(e)
−t
1
/RC
\begin{gathered}:\longmapsto \sf \cancel - \dfrac{1}{2} \log(2) = \cancel - \dfrac{t_1}{RC} \log \: e \\ \end{gathered}
:⟼
−
2
1
log(2)=
−
RC
t
1
loge
\begin{gathered}:\longmapsto \sf\dfrac{1}{2} \log2 = \dfrac{t_1}{RC} \\ \end{gathered}
:⟼
2
1
log2=
RC
t
1
:\longmapsto \boxed{ \pink{ \boxed{\bf t_2 = \frac{1}{2} \times RC \times log2}}}:⟼
t
2
=
2
1
×RC×log2
Hence,
\begin{gathered} \sf \dfrac{t_1}{t_2} = \frac{\frac{1}{2} \times \cancel{RC} \times \cancel{\log2}} { 2 \times \cancel{RC} \times \cancel{\log2}} \\ \end{gathered}
t
2
t
1
=
2×
RC
×
log2
2
1
×
RC
×
log2
\purple{ \large :\longmapsto \underline {\boxed{{\bf \frac{t_1}{t_2} = \frac{1}{4} } }}}:⟼
t
2
t
1
=
4
1
Step-by-step explanation:
Hope it helps